Your program ('nuggets') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

Your program ('heritage') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

刘汝佳：

Your program ('rockers') produced all correct answers!  This is your
submission #4 for this problem.  Congratulations!

哥的搜索，最慢的0.05
起初每什么思路，看别人的搜索程序看的难受，只好自己想写一个。
花了一些时间，没写过带这么多参数的搜索。

思想：对于每首歌只有刻CD，或者不刻CD，只记录当前状态，对于刻CD的情况，
1，如果当前CD的容量能装下就放在当前的CD里，并更新当前CD的容量vol。
2，如果不能放下，就放在下一张CD里，更新CD容量vol。(有点贪心的味道)
O(2^20)==1024*1024~~10^6

Your program ('fence9') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

nocow 皮克定理
a=S+1-b/2;   a是内部点的数量，S是面积，b是边上的点，
而(0,0)到(m,n)的边上的点为二者的最大公约数。

Your program ('camelot') produced all correct answers!  This is your
submission #11 for this problem.  Congratulations!

很麻烦的一个题目，最后一组数据打过去的，待修改。
用BFS求最短路径，用四维数组存所有的结果。载国王的情况只枚举了国王身边的八个点加上国王的位置（+-1），
即骑士先到这个点国王也到这个点，然后他们一起走。这可能就是最后一组过不去的原因，以前只有19组测试数据，最后一个可能官方后来加的，我同学国王身边+-2过了。
8 8 
D 5 
B 1 
F 1 
B 3

YOUR PROGRAM ('range') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

越来越感觉DP强大了：
dp[i][j]表示(i,j)为左上角坐标的时候可形成正方形的最大的边
如果dp[i][j]==0，dp[i][j]=0
否则，dp[i][j]=min(dp[i+1][j],dp[i][j+1],dp[i+1][j+1])+1
也就是说取(i,j)的右边，下边，右下边三者最小的进行扩展。
然后从右往左，从下到上进行dp即可。

取有两种情况

1，取i，留下区间(i+1,j)给对手，那么对手的得分是dp[i+1][j]

2,取j，留下区间(i,j-1)给对手，那么对手的得分是dp[i][j-1]

自己的得分是sum[i][j]-对手得分，于是dp[i][j]=sum[i][j]-min(dp[i+1][j],dp[i][j-1]);

USER: tian tianbing [tbbd4261]

TASK: fenceLANG: C++ Compiling...Compile: OKExecuting...

Test 1: TEST OK [0.022 secs, 4016 KB] 

Test 2: TEST OK [0.000 secs, 4016 KB]

Test 3: TEST OK [0.022 secs, 4016 KB] 

Test 4: TEST OK [0.022 secs, 4016 KB]

Test 5: TEST OK [0.011 secs, 4016 KB] 

Test 6: TEST OK [0.011 secs, 4016 KB]

Test 7: TEST OK [0.000 secs, 4016 KB] 

Test 8: TEST OK [0.043 secs, 4016 KB]All tests OK. 

Your program ('fence') produced all correct answers! 

This is your submission #3 for this problem. Congratulations!

欧拉回路也可能是欧拉路径，用dfs，从小到大搜，要求是输出字典序小的.

刚开始是输出放在函数里面，这样后两组测试数据过不了，

/*
ID:tbbd4261
PROG:fence
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("fence.in");
ofstream fout("fence.out");

const int MAX=505;
int adj[MAX][MAX]={0},F,nmax=0;
int cnt[MAX]={0};
int print[1025*2+100]={0},len=0;
void dfs(int u)
{
    //fout<<u<<endl;
    for(int v=1; v<=nmax; v++)
            if(adj[u][v])
            {
                         adj[u][v]--;
                         adj[v][u]--;
                         dfs(v);
            }     
    len++;
    print[len]=u;
}

int main()
{
    fin>>F;
    int i,s,t;
    for( i=1,s,t; i<=F; i++) 
     {
            fin>>s>>t;
            cnt[s]++;
            cnt[t]++;
            adj[s][t]++; 
            adj[t][s]++;
            if(s>nmax)nmax=s;
            if(t>nmax)nmax=t;
     }
     
     s=0;
     for(i=1; i<=nmax; i++)
              if(cnt[i]){
                         if(s==0)s=i;
                         if(cnt[i]%2==1){ s=i; break;}
              }
     
     dfs(s);
     for(int j=len; j>=1; j--)
             fout<<print[j]<<endl;
     
    return 0;
}

USER: tian tianbing [tbbd4261]
TASK: ratios
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.043 secs, 3016 KB]
Test 2: TEST OK [0.032 secs, 3016 KB]
Test 3: TEST OK [0.032 secs, 3016 KB]
Test 4: TEST OK [0.022 secs, 3016 KB]
Test 5: TEST OK [0.022 secs, 3016 KB]
Test 6: TEST OK [0.032 secs, 3016 KB]
All tests OK.
Your program ('ratios') produced all correct answers!  This is your
submission #18 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
5 8 0
3 5 4
1 3 0
6 2 0
------- test 2 -------
3 4 5
1 2 3
3 7 1
2 1 2
------- test 3 -------
20 22 24
4 5 5
9 5 6
12 12 14
------- test 4 -------
80 85 80
98 97 96
89 87 88
78 79 80
------- test 5 -------
99 99 99
1 0 0
0 1 0
0 0 1
------- test 6 -------
93 95 97
98 0 0
0 98 0
0 0 98

Thanks for your submission!

直接枚举就能过了，提交了18次，不知道为什么用fin读不了，后来还是让同学帮我改成C的文件读入才过的

到现在还是不知道为什么。

要注意比例和饲料都是非负的小于100的，也就是说可能出现0的情况，要小心。

/*
ID:tbbd4261
PROG:ratios
LANG:C++
*/

#include<fstream>
#include <stdio.h>
using namespace std;
//fstream fin("ratios.in");
ofstream fout("ratios.out");

int main()
{
    int food[4][4];
    freopen("ratios.in","r",stdin);
    freopen("ratios.out","w",stdout);
    int i,j,k,minsum=0xffffff,a=0,b=0,c=0,d=0;
    for(i=0; i<4; i++)
        for(j=1; j<=3; j++)
             scanf("%d",&food[i][j]);
    int n1,n2,n3;
    bool f1,f2,f=false;
    for(i=0; i<=100; i++)
        for(j=0; j<=100; j++)
             for(k=0; k<=100; k++){
                     n1=food[1][1]*i+food[2][1]*j+food[3][1]*k;
                     n2=food[1][2]*i+food[2][2]*j+food[3][2]*k;
                     n3=food[1][3]*i+food[2][3]*j+food[3][3]*k;
            
                 int t=0;  f1=false; f2=false;
                 if(food[0][1]==0){
                     f1=true;
                     if(n1!=0)continue;
                 }else{
                      if(n1%food[0][1]!=0)continue;
                      t=n1/food[0][1];
                 }
                 
                 if(food[0][2]==0){
                      f2=true;
                      if(n2!=0)continue;
                 }else {
                       if(n2%food[0][2]!=0)continue;
                       if(f1==true)t=n2/food[0][2];
                       else if(n2/food[0][2]!=t)continue;
                 }
                 
                 if(food[0][3]==0){
                       if(n3!=0)continue;  
                 }else{
                       if(n3%food[0][3]!=0)continue;
                       if(f1&&f2)t=n3/food[0][3];
                       if((f1==false ||f2==false)&&n3/food[0][3]!=t)continue;
                 }
             
                 if( n1+n2+n3<minsum&&(n1||n2||n3)){
                     f=true;
                  minsum=n1+n2+n3;  
                  a=i; b=j; c=k; d=t; 
                  }
    }
    if(f) fout<<a<<' '<<b<<' '<<c<<' '<<d<<endl;
    else fout<<"NONE"<<endl;   
    return 0;
}

USER: tian tianbing [tbbd4261]
TASK: spin
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 3012 KB]
Test 2: TEST OK [0.000 secs, 3012 KB]
Test 3: TEST OK [0.000 secs, 3012 KB]
Test 4: TEST OK [0.000 secs, 3012 KB]
Test 5: TEST OK [0.011 secs, 3012 KB]
Test 6: TEST OK [0.000 secs, 3012 KB]
Test 7: TEST OK [0.011 secs, 3012 KB]
Test 8: TEST OK [0.011 secs, 3012 KB]
All tests OK.
Your program ('spin') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
30 1 0 120
180 1 10 100
35 1 20 90
31 1 30 80
32 1 50 60
------- test 2 -------
30 1 350 350
180 1 10 100
35 1 67 23
31 1 30 4
32 1 50 7
------- test 3 -------
180 1 0 120
180 1 120 120
180 1 240 120
31 1 30 4
32 1 50 7
------- test 4 -------
1 1 140 359
1 1 200 359
1 1 4 1
2 1 6 1
1 1 300 300
------- test 5 -------
45 5 140 13 300 17 0 15 20 3 40 1
73 1 200 359
105 2 4 1 50 50
179 3 6 1 8 1 359 3
3 1 300 300
------- test 6 -------
45 5 120 13 200 17 0 15 20 3 32 1
73 5 200 100 1 30 50 10 70 2 75 1
105 2 100 1 50 20
179 3 6 1 8 1 359 3
3 1 300 359
------- test 7 -------
73 5 207 101 1 35 57 11 71 2 75 1
45 5 125 13 200 17 0 15 20 3 32 1
96 5 100 12 50 13 0 2 300 39 250 1
119 5 6 1 8 1 359 1 245 1 123 1
15 5 300 1 231 1 185 1 179 2 0 18
------- test 8 -------
73 5 207 11 1 35 57 11 71 2 75 1
45 5 125 3 200 17 0 15 20 3 32 1
96 5 100 1 50 1 0 2 300 39 250 1
119 5 6 1 8 1 359 1 245 1 123 1
15 5 300 1 231 1 185 1 179 2 0 1

Thanks for your submission!

直接枚举，t==360时所有轮子都回到最初状态。

用一个数组计数，f[i]==5的话说明所有轮子的缺口都经过这个地方，OKay！

/*
ID:tbbd4261
PROG:spin
LANG:C++
*/

#include<fstream>
#include<cstring>
using namespace std;
ifstream fin("spin.in");
ofstream fout("spin.out");
struct wheel
{
    int v,w;  //速度，缺口数 
    int s[5]; //缺口起始位置 
    int len[5]; //缺口大小 
}wheel[5];

int f[360]={0};
int main()
{
    for(int i=0; i<5; i++)
        {
            fin>>wheel[i].v>>wheel[i].w;
            for(int j=0; j<wheel[i].w; j++)
                fin>>wheel[i].s[j]>>wheel[i].len[j]; 
        }      
    int t;
    bool isP=false;
    for( t=0; t<360; t++ )
    {
         memset(f,0,sizeof f);
         for(int i=0 ; i<5; i++)
         {
               for(int j=0; j<wheel[i].w; j++ )
               {
                       int start= (wheel[i].s[j]+(wheel[i].v*t) )%360;
                       for(int k=0; k<=wheel[i].len[j]; k++ )
                       {
                             f[(k+start)%360]++;
                             if(f[(k+start)%360]>=5){ isP=true; fout<<t<<endl; return 0;}   
                       }
               }  
         }
             
    }
   fout<<"none"<<endl;
   
    return 0;
}

YOUR PROGRAM ('kimbits') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.

Here are the test data

/*
ID:tbbd4261
PROG:kimbits 
LANG:C++
*/

#include<fstream>
#include<string>
using namespace std;
ifstream fin("kimbits.in");
ofstream fout("kimbits.out");

int main()
{
    unsigned int cmb[32][32]={0};
    for(int i=0; i<32; i++)
            cmb[i][0]=1;
            
    for(int i=1; i<32; i++)
        for(int j=1; j<=i; j++ )
             cmb[i][j]=cmb[i-1][j-1]+cmb[i-1][j];
             
    for(int i=0; i<32; i++)
        for(int j=1; j<32; j++)
        cmb[i][j]+=cmb[i][j-1];
        
    unsigned n,l,i;
    fin>>n>>l>>i;
    for(unsigned ix=n; ix>0; --ix ){
         if(i>cmb[ix-1][l]){
              fout<<1;
              i-=cmb[ix-1][l];
              l--;
         }
         else fout<<0;
    }  
    fout<<endl;
    return 0;
}

USER: tian tianbing [tbbd4261]
TASK: fact4
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 3028 KB]
Test 2: TEST OK [0.011 secs, 3028 KB]
Test 3: TEST OK [0.000 secs, 3028 KB]
Test 4: TEST OK [0.011 secs, 3028 KB]
Test 5: TEST OK [0.000 secs, 3028 KB]
Test 6: TEST OK [0.011 secs, 3028 KB]
Test 7: TEST OK [0.000 secs, 3028 KB]
Test 8: TEST OK [0.000 secs, 3028 KB]
Test 9: TEST OK [0.000 secs, 3028 KB]
Test 10: TEST OK [0.011 secs, 3028 KB]
All tests OK.
Your program ('fact4') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
1
------- test 2 -------
2
------- test 3 -------
7
------- test 4 -------
14
------- test 5 -------
143
------- test 6 -------
645
------- test 7 -------
777
------- test 8 -------
888
------- test 9 -------
999
------- test 10 -------
1000

/*
ID:tbbd4261
PROG:fact4
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("fact4.in");
ofstream fout("fact4.out");
int dp[4225]={0},n,i;
int cal(int a, int b)
{
     int temp=a*b;
     while(temp%10==0)
     {
                temp/=10;
     }
     return temp%100000;
}
int main()
{
    dp[1]=1;
    fin>>n;
    for(i=2; i<=n; i++)
    {
        dp[i]=cal(i,dp[i-1]);        
    }
    fout<<dp[n]%10<<endl;
    
    return 0;
}

10=2*5;

统计N！中2的个数和5的个数，用digit记录除去2和5的最后一位阶乘结果。

最后把多余的2的个数乘回去（即n2-n5个2）

/*
ID:tbbd4261
PROG:fact4
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("fact4.in");
ofstream fout("fact4.out");
int main()
{
    int n,i,digit,n2,n5,j;
    fin>>n;
    n2=n5=0;
    digit=1;
    for(i=2; i<=n; i++){
        j=i;
        while(j%2==0){
             n2++;
             j/=2;
        }
        
        while(j%5==0){
              n5++;
              j/=5; 
        }
        digit=(digit*j)%10;
    }
    for(i=0; i<n2-n5; i++)
             digit=(digit*2)%10;
    fout<<digit%10<<endl;
    return 0;
}

USER: tian tianbing [tbbd4261]
TASK: stamps
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 10864 KB]
Test 2: TEST OK [0.000 secs, 10864 KB]
Test 3: TEST OK [0.000 secs, 10864 KB]
Test 4: TEST OK [0.011 secs, 10864 KB]
Test 5: TEST OK [0.000 secs, 10864 KB]
Test 6: TEST OK [0.011 secs, 10864 KB]
Test 7: TEST OK [0.011 secs, 10864 KB]
Test 8: TEST OK [0.011 secs, 10864 KB]
Test 9: TEST OK [0.000 secs, 10864 KB]
Test 10: TEST OK [0.162 secs, 10864 KB]
Test 11: TEST OK [0.983 secs, 10864 KB]
Test 12: TEST OK [0.324 secs, 10864 KB]
Test 13: TEST OK [0.011 secs, 10864 KB]
All tests OK.

Your program ('stamps') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

DP：

 f[i]表示凑成i分游子需要的最少邮票数。
 f[i]=min{f[i-stamp[j]]+1}，(j=1……n)&&i>=stamp[j];

 初始状态f[0]=0;

/*
ID:tbbd4261
PROG:stamps
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("stamps.in");
ofstream fout("stamps.out");
const int INTMAX=0x0fffffff;
int n,k,i,j,stamp[55],minn;
int f[201*10001]={0};
int main()
{
    fin>>k>>n;
    for(i=1; i<=n; i++)
             fin>>stamp[i];
    f[0]=0;
    for(i=1; i<=200*10000; i++)
    {
             
             f[i]=INTMAX; 
             for(j=1; j<=n; j++)
             {
                  if(i>=stamp[j]&&f[i]>f[i-stamp[j]]+1)
                       f[i]=f[i-stamp[j]]+1;         
             }
             if(f[i]==INTMAX||f[i]>k)break;
    }
    fout<<i-1<<endl;
    return 0;
}

Your program ('contact') produced all correct answers!  This is your
submission #7 for this problem.  Congratulations!

交了7次，被格式给搞败掉了。

还有就是strig的构造函数，string str(s+i,j);str为s+i位置开始的j个字符，而不是string str(s,i,j);

定义一个map<string ,int>用于给各个串计数，然后转到vector中排序。 还好能过。

/*
ID:tbbd4261
PROG:contact
LANG:C++
*/

#include<fstream>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
ifstream fin("contact.in");
ofstream fout("contact.out");
int a, b ,n ,i,j,cnt=0 ;
char s[200005]={0};
bool cmp(const pair<int,string> &a, const pair<int,string > &b )
{
     if(a.first>b.first)return true;
     else if(a.first<b.first)return false;
     if(a.second.size()<b.second.size())return true;
     if(a.second.size()>b.second.size())return false;
     return a.second<b.second;
}

int main()
{  
    char ch;
    map<string,int> mp;
    fin>>a>>b>>n;
    
    for(i=1;fin>>ch; i++)
            s[i]=ch;
    cnt=--i;
    for(i=a; i<=b; i++ )
    {
             for(j=1;j+i-1<=cnt;j++)
             {
                  string str(s+j,i);//错的地方 
                  mp[str]++;
             }
    }
    
    vector< pair<int,string> > vec;
    map<string ,int>::iterator it=mp.begin();
    while(it!=mp.end())
    {
          vec.push_back(make_pair(it->second,it->first));
          it++;
    }
    
    sort(vec.begin(),vec.end(),cmp);
    int len,j=0,c;
    for(i=1; i<=n&&j!=vec.size(); i++)
    {
             len=vec[j].first;
             fout<<len<<endl;  c=0;
             while(j!=vec.size()&&vec[j].first==len)
             {
                    
                    if(c&&c%6!=0)fout<<' ';
                    fout<<vec[j].second;
                    c++;
                    if(c%6==0)fout<<endl;
                    j++;
             }                    
             if(c%6!=0)fout<<endl;
    }
    
    return 0;
}

Your program ('rect1') produced all correct answers!  This is your
submission #15 for this problem.  Congratulations!

     终于做出来了，用队列实现，没接收一个就遍历整个队列。要注意的是要把队列的大小用t记下，因为

新加进来的会改变队列的大小。

    用的是矩形分割：

原来的程序在分割与a相交的矩形b的时候，把a与b相交的部分也作为一个小矩形放在了后面。

这也就会多出很多，统计了一下对于最后一组测试数据有40多万个矩形，超时。

其实应该把a整个放进去，不该把a也分割的。

YOUR PROGRAM ('humble') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated

USACO上给了一个高效数据结构也不复杂的处理方法。

for each prime p
find the minimum humble number h
such that h * p is bigger than the last humble number.
take the smallest h * p found: that's the next humble number.

假设前面已经确定k个丑数。

      对于每个质数p（从前往后遍历，由题意知是递增的）

             找到最小的丑数h，使 h*p > humble[k]（不断用这个更新min）

 min就是下一个要求的丑数humble[k+1];

    技巧的地方：pindex[i]数组记录记录第i个质数（data[i]）乘到（humble[]表）的地方，

因为以后的数一定比这个大，不需要从头再扫。

/*
ID:tbbd4261
PROG:humble
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("humble.in");
ofstream fout("humble.out");
const int MAX=100005;
long long data[105]={0}, humble[MAX]={0},pindex[105]={0};
int n,k,i,nhum=0;
int main()
{
    fin>>k>>n;
    for(i=1; i<=k; i++)
        fin>>data[i];
    
    humble[0]=1;
    humble[1]=data[1];
    nhum=1;
    for(i=1; i<=k; i++)pindex[i]=0;
    while(nhum<=n)
    {
                 
                  int min=0x7fffffff,temp,tt=1;
                  for(i=1; i<=k; i++)
                  {
                           while(data[i] * humble[pindex[i]]<=humble[nhum])
                                          pindex[i]++;
                           temp=data[i]*humble[ pindex[i] ];
                           if( temp<min){ min=temp; tt=i; }
                          
                  }
                  humble[++nhum]=min;
                  pindex[tt]++;
    }
    
    fout<<humble[n]<<endl;
 
    return 0;
}

YOUR PROGRAM ('inflate') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

/*
ID:tbbd4261
PROG:inflate
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("inflate.in");
ofstream fout("inflate.out");
const int MAX=10005;
int f[MAX]={0};
int p[MAX],t[MAX];
int main()
{
    int n,m,i,j;
    fin>>m>>n;
    for(i=1; i<=n; i++)
        fin>>p[i]>>t[i];
    
    for(i=1; i<=n; i++)
    for(j=t[i]; j<=m; j++)
    {
             f[j]=(f[j-t[i]]+p[i]>f[j]?f[j-t[i]]+p[i]:f[j] );
    }
    fout<<f[m]<<endl;
    
    return  0;
}
 


     

Your program ('agrinet') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!
最小生成树：

/*
ID:tbbd4261
PROG:agrinet
LANG:C++
*/

#include<fstream>
#include<string.h>
#include<cmath>
#include<algorithm>
using namespace std;
ifstream fin("agrinet.in");
ofstream fout("agrinet.out");

int fa[10100];   int u[10101]; int v[10101]; 
int w[10101];    int r[10101];
bool cmp(int i,int j){return w[i]<w[j]; }
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]); }
int main()
{
    int n,i,j,dis,count=0,ans=0;
    fin>>n;
    for(i=1; i<=n; i++)
    for(j=1; j<=n; j++)
    {
      fin>>dis;
      if(dis){ u[++count]=i; v[count]=j; w[count]=dis; }
    } 
    for(i=1; i<=count; i++)r[i]=i;
    for(i=1; i<=count; i++)fa[i]=i;
    
    sort(r+1,r+1+count,cmp);
    for(i=1; i<=count; i++)
    {
     int e=r[i]; int x=find(u[e]); int y=find(v[e]);
     if(x!=y){ ans+=w[e]; fa[x]=y; } 
    }
    fout<<ans<<endl;
    return 0;
}

USER: tian tianbing [tbbd4261]
TASK: fracdec
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 3792 KB]
Test 2: TEST OK [0.000 secs, 3792 KB]
Test 3: TEST OK [0.000 secs, 3792 KB]
Test 4: TEST OK [0.000 secs, 3792 KB]
Test 5: TEST OK [0.000 secs, 3792 KB]
Test 6: TEST OK [0.000 secs, 3792 KB]
Test 7: TEST OK [0.022 secs, 3792 KB]
Test 8: TEST OK [0.032 secs, 3792 KB]
Test 9: TEST OK [0.000 secs, 3792 KB]
All tests OK.
Your program ('fracdec') produced all correct answers!  This is your
submission #4 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
22 5
------- test 2 -------
1 7
------- test 3 -------
100000 59
------- test 4 -------
1 100000
------- test 5 -------
3 3
------- test 6 -------
59 330
------- test 7 -------
100000 9817
------- test 8 -------
1 99991
------- test 9 -------
982 4885

/*
ID:tbbd4261
PROG:fracdec
LANG:C++
*/

#include<fstream>
#include<string.h>
#include<cmath>
using namespace std;
ifstream fin("fracdec.in");
ofstream fout("fracdec.out");
const int MAX=100005;
int n,d,i,temp,len=0,j;
int k[MAX]={0}, dat[MAX]={0};
int gcd(int x, int y)
{
    return x==0?y:gcd( y%x,x );
}

int main()
{
    fin>>n>>d;
    fout<<n/d<<'.';
    if(n>=d)len=int(log(n/d)/log(10)+2.0001);
    else len=2;
    n%=d;
    int temp=d/gcd(n,d);
    
    while(temp%2==0)temp/=2;
    while(temp%5==0)temp/=5;
    if(temp==1)//无循环 
    {
               while(1)
               {
                       
                       n*=10;
                       fout<<n/d;
                       n%=d;
                       len++;
                       if(n==0)break;
               }
               fout<<endl;
               return 0;
    }
    for(i=1;;i++)
    {
           if(k[n]!=0)
           {
                for(j=1; j<=k[n]-1; j++)
                {
                         fout<<dat[j];
                         len++;
                         if(len%76==0)fout<<endl;
                }
                fout<<'(';
                len++;
                if(len%76==0)fout<<endl;
                for(j=k[n]; j<i; j++)
                {
                            fout<<dat[j];
                            len++;
                            if(len%76==0)fout<<endl;
                }
                fout<<')'<<endl;
                return 0;
                      
           }
           k[n]=i;
           n*=10;
           dat[i]=n/d;
           n%=d;
    }
    return 0;
}

YOUR PROGRAM ('comehome') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.
学校OJ代码：

Floyd:

/*
ID:tbbd4261
PROG:comehome
LANG:C++
*/

#include<fstream>
#include<string.h>
#include<iostream>
using namespace std;
ifstream fin("comehome.in");
ofstream fout("comehome.out");
int dis[53][53];
const int INF=10000000;
void Floyd(int n)
{
     for(int k=1; k<=n; k++)
     for(int i=1; i<=n; i++)
     for(int j=1; j<=n; j++)
      if(i!=k&&k!=j&&i!=j&&dis[i][k]+dis[k][j]<dis[i][j])
      dis[i][j]=dis[i][k]+dis[k][j];

}

int main()
{
  int p,i,j,k,d,n1,n2; 
    fin>>p;
    memset(dis,0,sizeof (dis));
    for(i=1; i<=52; i++)
    for(j=1; j<=52; j++)
    dis[i][j]=INF;
     
    for(i=1; i<=p; i++)
    {
       char v1,v2;
       fin>>v1>>v2>>d;      
       if(v1==v2)continue;
       n1=(v1>='a'?v1-'a'+1:v1-'A'+26+1);
       n2=(v2>='a'?v2-'a'+1:v2-'A'+26+1);
       if(d<dis[n1][n2])dis[n1][n2]=dis[n2][n1]=d;
    }
    
    Floyd(52);
    
    int min=INF+100;
    char c;
    for(i=27; i<=51; i++) //大写字母到Z 
    {
       if(dis[i][52]<min){min=dis[i][52];c=i; }
    }
    fout<<char(c-27+'A')<<' '<<min<<endl;

  
    return 0;
}

Djkstra:

/*
ID:tbbd4261
PROG:comehome
LANG:C++
*/

#include<fstream>
#include<string.h>
using namespace std;
ifstream fin("comehome.in");
ofstream fout("comehome.out");
int adj[53][53];
int dis[53];
bool f[53];
const int INF=10000000;
int main()
{
  int p,i,j,k,d,n1,n2; 
    fin>>p;
    memset(f,0,sizeof (f));
    for(i=1; i<=52; i++)
    for(j=1; j<=52; j++)
           adj[i][j]=INF;
    for(i=1; i<=p; i++)
    {
       char v1,v2;
       fin>>v1>>v2>>d;      
       if(v1==v2)continue;
       n1=(v1>='a'?v1-'a'+1:v1-'A'+26+1);
       n2=(v2>='a'?v2-'a'+1:v2-'A'+26+1);
       if(d<adj[n1][n2])
       adj[n1][n2]=adj[n2][n1]=d;
    }
    
    
    for(i=1; i<=51; i++)
        if(adj[52][i])dis[i]=adj[52][i];
        else dis[i]=INF;     
    
    f[52]=1;  dis[52]=0; int min,u;
    for(i=1; i<=52; i++)
    {       min=INF; u=0;
            for(j=1; j<=52; j++)
                     if(!f[j]&&dis[j]<min){ min=dis[j]; u=j; }
            f[u]=1; 
            if(u)
            for(j=1; j<=52; j++)
                if(!f[j]&&dis[j]>dis[u]+adj[u][j])dis[j]=dis[u]+adj[u][j];
    }
    min=INF;  int c;
    for(i=27; i<=51; i++) //大写字母到Z 
    {
       if(dis[i]<min){min=dis[i]; c=i; }
    }
    fout<<char(c-27+'A')<<' '<<min<<endl;
  
    return 0;
}

Your program ('cowtour') produced all correct answers! This is your submission #2 for this problem. Congratulations!

Your program ('maze1') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

DFS从两个入口各搜索一次，更新即可。

先写了一个代码，太乱，花了不少时间，后来重写了，要注意的代码的整洁。

/*
ID:tbbd4261
PROG:maze1
LANG:C++
*/

#include<fstream>
using namespace std;
ifstream fin("maze1.in");
ofstream fout("maze1.out");
char wall[205][80]={0};
int step[105][40]={0};
char ch;
int x1=0,y1=0,x2=0,y2=0,i,j,w,h;
int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};


void input()
{
     fin>>w>>h;  
     fin.get();
     for(i=1; i<=2*h+1; i++)
     {
        for(j=1; j<=2*w+1; j++)
              {
                   fin.get(ch); wall[i][j]=ch;
                   if((i==1||i==2*h+1||j==1||j==2*w+1)&&ch==' ')
                   {
                             if(!x1){ x1=i; y1=j; }
                             else {x2=i; y2=j; }
                   } 
              }
        fin.get();
     }
}


bool valid(int i, int j)
{ return i>=2&&i<=2*h&&j>=2&&j<=2*w&&(i%2==0)&&(j%2==0) ; }

bool valid2(int i, int j){return i>=1&&i<=h&&j>=1&&j<=w; }

void deal(int &x, int &y)
{
     for( i=0; i<4; i++ )
         if(valid(x+dx[i],y+dy[i]))
         { x=x+dx[i]; y=y+dy[i]; }
     x/=2; y/=2;
}

void dfs(int i, int j)
{
      for(int k=0; k<4; k++)
          if(valid2(i+dx[k],j+dy[k])&&wall[2*i+dx[k]][j*2+dy[k]]==' ')
               if((step[i+dx[k]][j+dy[k]]>step[i][j]+1)||step[i+dx[k]][j+dy[k]]==0)      
               {  step[i+dx[k]][j+dy[k]]=step[i][j]+1; dfs(i+dx[k],j+dy[k]); }
}
int main()
{
    input();
    deal(x1,y1);
    deal(x2,y2);
    step[x1][y1]=1;
    step[x2][y2]=1;
    dfs(x1,y1);
    dfs(x2,y2);
    int max=1;
    for(i=1; i<=h; i++)
    for(j=1; j<=w; j++)
       if(step[i][j]>max)max=step[i][j];
    fout<<max<<endl;   
    return 0;
}