C++博客-烟雨江南-随笔分类-USACO 题解

USACO-The Clocks

laysent — Sun, 15 Nov 2009 03:49:00 GMT

一开始想用DFS解决，结果发现实在想不出什么好一点的启发函数可以让搜索较快得到优先解。然后就改用ID，但是时间拖得太长了，只过了两个数据就卡住了。最后想来想去想不好，干脆用穷举法，反正才4^9种可能。终于过了。

#include"stdio.h"
FILE *fin,*fout;
int answer[33],a[9],maxcount,mincount=28,sum[10];

int change(int a)
{
    if (a<12)
        return (a+3);
    if (a==12)
        return (3);
}

search(s1,s2,s3,s4,s5,s6,s7,s8,s9)
int s1,s2,s3,s4,s5,s6,s7,s8,s9;
{
    int k,j,count=0,check=0,s[10],aa,b,c,d,e,f,h,g,i;
    s[1]=s1;s[2]=s2;s[3]=s3;s[4]=s4;s[5]=s5;s[6]=s6;s[7]=s7;s[8]=s8;s[9]=s9;
    aa=a[0];b=a[1];c=a[2];d=a[3];e=a[4];f=a[5];g=a[6];h=a[7];i=a[8];
    for (k=1;k<=9;k++)
    {
        for (j=1;j<=s[k];j++)/*改变clocks的数值*/
        {
            if (k==1) {aa=change(aa);b=change(b);d=change(d);e=change(e);}
            if (k==2) {aa=change(aa);b=change(b);c=change(c);}
            if (k==3) {b=change(b);c=change(c);e=change(e);f=change(f);}
            if (k==4) {aa=change(aa);d=change(d);g=change(g);}
            if (k==5) {b=change(b);d=change(d);e=change(e);f=change(f);h=change(h);}
            if (k==6) {c=change(c);f=change(f);i=change(i);}
            if (k==7) {d=change(d);e=change(e);g=change(g);h=change(h);}
            if (k==8) {g=change(g);h=change(h);i=change(i);}
            if (k==9) {e=change(e);f=change(f);h=change(h);i=change(i);}
        }
    }
    check=0;
    if (aa==12 && b==12 && c==12 && d==12 && e==12 && f==12 && g==12 && h==12 && i==12)/*验证是否满足*/
        check=1;
    count=0;
    if (check==1) /*满足则记录答案*/
    {
        for (k=1;k<=9;k++)
        {
            for (j=1;j<=s[k];j++)
            {
                answer[count]=k;
                count++;
            }
        }
        mincount=count-1;
    }

}

main()
{
    int i,j,tem,s1,s2,s3,s4,s5,s6,s7,s8,s9;
    fin=fopen("clocks.in","r");
    for (i=0;i<7;i+=3)
        fscanf(fin,"%d %d %d\n",&a[i],&a[i+1],&a[i+2]);
    fclose(fin);
    for (i=0;i<=10;i++)
        sum[i]=0;
    for (s9=0;s9<=3;s9++)
        for (s8=0;s8<=3;s8++)
            for (s7=0;s7<=3;s7++)
                for (s6=0;s6<=3;s6++)
                    for (s5=0;s5<=3;s5++)
                        for (s4=0;s4<=3;s4++)
                            for (s3=0;s3<=3;s3++)
                                for (s2=0;s2<=3;s2++)
                                    for (s1=0;s1<=3;s1++) /*遍历所有可能的情况*/
                                    {
                                        if (s1+s2+s3+s4+s5+s6+s7+s8+s9                                        search(s1,s2,s3,s4,s5,s6,s7,s8,s9);
                                    }
    for (i=0;i<=mincount;i++)
        for (j=i+1;j<=mincount;j++) /*将答案自小到大排列*/
            if (answer[i]>answer[j])
            {
                tem=answer[i];
                answer[i]=answer[j];
                answer[j]=tem;
            }
    fout=fopen("clocks.out","w");
    fprintf(fout,"%d",answer[0]);
    for (i=1;i<=mincount;i++)
        fprintf(fout," %d",answer[i]);
    fprintf(fout,"\n");
    fclose(fout);
    return 0;
}

laysent 2009-11-15 11:49 发表评论

USACO-Packing Rectangles

laysent — Sat, 14 Nov 2009 09:29:00 GMT

不太简单的一道题目。如果USACO没有给出提示的话我想我是做不出来的。
一共有六种情况，其中有两种是相似的，可以归并为一种。有了提示，方法也就容易想到了。就是列举所有的可能性，然后放到五种情况里面去计算面积大小。只不过写起来就比较麻烦了。

#include "stdio.h"
#define MAX 10000 /*MAX记录面积最大上限，用于初始化*/
FILE *fin,*fout;
int maxs=MAX,a[3][10], answer[3][10],count=0;/*maxs记录最大面积，a存储矩形边长，count用于计算答案个数*/

int change(int i)
{
    if (i==1)
        return(2);
    if(i==2)
        return(1);
}

int max(a,b,c,d)/*找最大值*/
int a,b,c,d;
{
    int maxn;
    maxn=a;
    if (maxn        maxn=b;
    if (maxn        maxn=c;
    if (maxn        maxn=d;
    return maxn;
}

void culculate(length,width) /*计算面积并记录*/
int length,width;
{
    int i,check=0;
    if (maxs==length*width)
    {
        check=0;
        for (i=0;i            if (answer[1][i]==(length>width?length:width))
                check=1;
        if (check==0)
        {answer[1][count]=(length>width?length:width);
        answer[2][count]=(length        count++;}
    }
    if (maxs>length*width)
    {
        maxs=length*width;
        count=0;
        answer[1][count]=(length>width?length:width);
        answer[2][count]=(length        count++;
    }
}

void simulate(s1,d1,s2,d2,s3,d3,s4,d4) /*模拟所有的可能性*/
int s1,d1,s2,d2,s3,d3,s4,d4;
{
    int k1,k2,k3,k4,length,width;
    k1=change(d1);k2=change(d2);k3=change(d3);k4=change(d4);
    /*模拟第一种情况*/
    length=a[d1][s1]+a[d2][s2]+a[d3][s3]+a[d4][s4];
    width=max(a[k1][s1],a[k2][s2],a[k3][s3],a[k4][s4]);
    culculate(length,width);
    /*模拟第二种情况*/
    length=max(a[d1][s1]+a[d2][s2]+a[d3][s3],a[d4][s4],0,0);
    width=max(a[k1][s1],a[k2][s2],a[k3][s3],0)+a[k4][s4];
    culculate(length,width);
    /*模拟第三种情况*/
    length=max(a[d1][s1]+a[d2][s2],a[d3][s3],0,0)+a[d4][s4];
    width=max(a[k1][s1]+a[k3][s3],a[k3][s3]+a[k2][s2],a[k4][s4],0);
    culculate(length,width);
    /*模拟第四种情况*/
    length=a[d1][s1]+a[d2][s2]+max(a[d3][s3],a[d4][s4],0,0);
    width=max(a[k1][s1],a[k2][s2],a[k3][s3]+a[k4][s4],0);
    culculate(length,width);
    /*模拟第五种情况*/
    if (a[k3][s3]>a[k2][s2]+a[k4][s4])
        length=max(a[d1][s1],a[d2][s2]+a[d3][s3],a[d3][s3]+a[d4][s4],0);
    if (a[k3][s3]>a[k4][s4] && a[k3][s3]        length=max(a[d1][s1]+a[d2][s2],a[d2][s2]+a[d3][s3],a[d3][s3]+a[d4][s4],0);
    if (a[k4][s4]>a[k3][s3] && a[k4][s4]        length=max(a[d1][s1]+a[d2][s2],a[d1][s1]+a[d4][s4],a[d3][s3]+a[d4][s4],0);
    if (a[k4][s4]>=a[k1][s1]+a[k3][s3])
        length=max(a[d2][s2],a[d1][s1]+a[d4][s4],a[d3][s3]+a[d4][s4],0);
    if (a[k3][s3]==a[k4][s4])
        length=max(a[d1][s1]+a[d2][s2],a[d3][s3]+a[d4][s4],0,0);
    width=max(a[k1][s1]+a[k3][s3],a[k2][s2]+a[k4][s4],0,0);
    culculate(length,width);
}

sort()
{
    int tem,i,j;
    for (i=0;i        for (j=i+1;j        {
            if (answer[2][i]>answer[2][j])
            {
                tem=answer[1][i];
                answer[1][i]=answer[1][j];
                answer[1][j]=tem;
                tem=answer[2][i];
                answer[2][i]=answer[2][j];
                answer[2][j]=tem;
            }
        }
    fout=fopen("packrec.out","w");
    fprintf(fout,"%d\n",maxs);
    tem=0;
    for (i=0;i    {
        if (answer[2][i]!=tem)
            fprintf(fout,"%d %d\n",answer[2][i],answer[1][i]);
        tem=answer[2][i];
    }
    fclose(fout);
}

main()
{
    int i,s1,s2,s3,s4,d1,d2,d3,d4;
    fin=fopen("packrec.in","r");
    for (i=1;i<5;i++)
        fscanf(fin,"%d %d\n",&a[1][i],&a[2][i]);
    fclose(fin); /*读入数据*/
    for (s1=1;s1<5;s1++) /*排列组合所有可能的情况*/
        {for (s2=1;s2<5;s2++)
            {if (s1!=s2)
                {for(s3=1;s3<5;s3++)
                {if (s3!=s2 && s3!=s1)
                    {for (s4=1;s4<5;s4++)
                    {if (s4!=s1 && s4!=s2&& s4!=s3)
                    {for (d1=1;d1<3;d1++)/*以下变量的变化决定了矩形是横的还是竖的*/
                        {for (d2=1;d2<3;d2++)
                            {for (d3=1;d3<3;d3++)
                                {for (d4=1;d4<3;d4++)
                                    simulate(s1,d1,s2,d2,s3,d3,s4,d4);}}}}}}}}}}
    sort();
    return 0;
}

laysent 2009-11-14 17:29 发表评论

USACO-Barn Repair

laysent — Mon, 09 Nov 2009 07:45:00 GMT

#include "stdio.h"
#define MAX 200
FILE *fin,*fout;

void sort(temp,start,end)
int temp[],start,end;
{
int i,j,tem;
for (i=start;i<=end;i++)
{
  for (j=i+1;j<=end;j++)
  {
   if (temp[i]>temp[j])
   {
    tem=temp[i];
    temp[i]=temp[j];
    temp[j]=tem;
   }
  }
}
}

main()
{
int i,stall,maxn,cown,cow[202],crack[203],start,count,lenth=0,need=0;
fin=fopen("barn1.in","r");
fscanf(fin,"%d %d %d\n",&stall,&maxn,&cown);
for (i=0;i  fscanf(fin,"%d\n",&cow[i]);
fclose(fin);
sort(cow,0,cown-1);
fout=fopen("barn1.out","w");
start=cow[0];
count=0;
for (i=1;i  {
   if (cow[i]-cow[i-1]!=1)
   {
    crack[count]=cow[i]-cow[i-1]-1;
    start=cow[i];
    count++;
   }
  }
lenth=cown;
need=count-stall+1;
sort(crack,0,count-1);
for (i=0;i  lenth+=crack[i];
fprintf(fout,"%d\n",lenth);
fclose(fout);
return 0;
}

laysent 2009-11-09 15:45 发表评论

USACO-Mixing Milk

laysent — Sat, 07 Nov 2009 05:33:00 GMT

#include "stdio.h"
#define MAX 5000/* the number of farmers at most*/
FILE *fin,*fout;
int cent[MAX];long amount[MAX];

void quicksort(int start,int end)
{
int i,j,temp;
long longtemp;
i=start;j=end;
while (i{
while (cent[i]<=cent[j] && i i++;
if (cent[i]>cent[j])
{temp=cent[j];
cent[j]=cent[i];
cent[i]=temp;
longtemp=amount[j];
amount[j]=amount[i];
amount[i]=longtemp;}
while (cent[j]>=cent[i] && i j--;
if (cent[j] {temp=cent[j];
cent[j]=cent[i];
cent[i]=temp;
longtemp=amount[j];
amount[j]=amount[i];
amount[i]=longtemp;}
}
if (startif (i}

main()
{
int num,i;
long aim,money=0,have=0;
fin=fopen("milk.in","r");
fscanf(fin,"%ld %d\n",&aim,&num);
for (i=0;i fscanf(fin,"%d %ld\n",¢[i],&amount[i]);
fclose(fin);
quicksort(0,num-1);
for (i=0;i{
if (have>=aim)
  break;
else
{
  if (have+amount[i]<=aim)
  {
   have+=amount[i];
   money+=cent[i]*amount[i];
  }
  else
  {
   money+=cent[i]*(aim-have);
   have=aim;
  }
}
}
fout=fopen("milk.out","w");
fprintf(fout,"%ld\n",money);
fclose(fout);
return (0);
}

laysent 2009-11-07 13:33 发表评论

USACO-Dual Palindromes

laysent — Wed, 04 Nov 2009 03:31:00 GMT

#include "stdio.h"
FILE *fin,*fout;
int counter;

palindrom(int zahl)
{int count=0,check=0,basis,quadrat,i,j,k;
char number[20];
for (basis=2;basis<=10;basis++)
{
quadrat=zahl;
for (i=0;i<=count;i++)
    number[i]='0';
number[0]=quadrat % basis +48;
count=1;
quadrat=quadrat/basis;
while (quadrat>=basis)
{
number[count]=quadrat % basis +48;
count++;quadrat=quadrat/basis;
}
if (quadrat!=0) {number[count]=quadrat+48;count++;}
k=1;
for(i=0,j=count-1;i<=j;i++,j--)
    if (number[i]!=number[j])
    {k=0;break;}
if (k==1)
check++;
if (check>=2)
{
    fprintf(fout,"%d\n",zahl);
    counter++;
    break;
}
}
}

main()
{
    int num,startnum,pronum;
    fin=fopen("dualpal.in","r");
    fout=fopen("dualpal.out","w");
    fscanf(fin,"%d %d\n",&num,&startnum);
    fclose(fin);
    counter=0;
    while (counter    {
        startnum++;
        pronum=startnum;
        palindrom(pronum);
    }
    return(0);
}

laysent 2009-11-04 11:31 发表评论

USACO-Palindromic Squares

laysent — Wed, 04 Nov 2009 03:04:00 GMT

这个程序自以为写得极差。因为一开始题目没看清楚的缘故，结果平白无故又只能多加几个函数上去。整个程序很冗长。真是败笔啊。

#include "stdio.h"
#define MAX 300
char number1[17],number2[17];
int base;
FILE *fin,*fout;

void transform(int stop)
{
    int i,j;
    char tem;
    for (i=0,j=stop-1;i<=j;i++,j--)
        {
            tem=number1[i];
            number1[i]=number1[j];
            number1[j]=tem;
        }
}

trans (quadrat,basis)
int quadrat,basis;
{int count;
if((quadrat %basis)> 9)
number1[0]=quadrat % basis+55;
else number1[0]=quadrat % basis +48;
count=1;
quadrat=quadrat/basis;
while (quadrat>=basis)
{ if((quadrat %basis)> 9)
number1[count]=quadrat % basis+55;
else number1[count]=quadrat % basis +48;
count++;quadrat=quadrat/basis;
}
if (quadrat>9)
number1[count]=quadrat+55;
else if (quadrat!=0) {number1[count]=quadrat+48;count++;
    transform(count);}
}

int transbase (quadrat,basis)
int quadrat,basis;
{int count;
if((quadrat %basis)> 9)
number2[0]=quadrat % basis+55;
else number2[0]=quadrat % basis +48;
count=1;
quadrat=quadrat/basis;
while (quadrat>=basis)
{ if((quadrat %basis)> 9)
number2[count]=quadrat % basis+55;
else number2[count]=quadrat % basis +48;
count++;quadrat=quadrat/basis;
}
if (quadrat>9)
{number2[count]=quadrat+55;count++;}
else if (quadrat!=0) {number2[count]=quadrat+48;count++;}
return(count);}

check(counter,ii)
int counter,ii;
{int i,j,k;
k=1;
for(i=0,j=counter-1;i<=j;i++,j--)
    if (number2[i]!=number2[j])
    k=0;
if (k==1)
{ trans(ii,base);
fprintf(fout,"%s %s\n",number1,number2);}}

main()
{
int i,trans,square;
fin=fopen("palsquare.in","r");
fout=fopen("palsquare.out","w");
fscanf(fin,"%d\n",&base);
for (i=1;i<=MAX;i++)
{square=i*i;
trans=transbase(square,base);
check(trans,i);}
return(0);
}

laysent 2009-11-04 11:04 发表评论

USACO-Name That Number

laysent — Sun, 01 Nov 2009 14:01:00 GMT

#include "stdio.h"
FILE *fin,*fout,*fdict;
char number[15],print[15];
static int checkstop[8]={67,70,73,76,79,83,86,89},checkbegin[8]={65,68,71,74,77,80,74,87};

int check(int counter)
{
    if (number[counter]=='2' && (print[counter]=='A' || print[counter]=='B'||print[counter]=='C')) return 1;
    if (number[counter]=='3' && (print[counter]=='D' || print[counter]=='E'||print[counter]=='F')) return 1;
    if (number[counter]=='4' && (print[counter]=='G' || print[counter]=='H'||print[counter]=='I')) return 1;
    if (number[counter]=='5' && (print[counter]=='J' || print[counter]=='K'||print[counter]=='L')) return 1;
    if (number[counter]=='6' && (print[counter]=='M' || print[counter]=='N'||print[counter]=='O')) return 1;
    if (number[counter]=='7' && (print[counter]=='P' || print[counter]=='R'||print[counter]=='S')) return 1;
    if (number[counter]=='8' && (print[counter]=='T' || print[counter]=='U'||print[counter]=='V')) return 1;
    if (number[counter]=='9' && (print[counter]=='W' || print[counter]=='X'||print[counter]=='Y')) return 1;
    return 0;
}

main()
{
    int num,count,printcheck,printcount=0,i,begin,stop;
    fin=fopen("namenum.in","r");
    fdict=fopen("dict.txt","r");
    fout=fopen("namenum.out","w");
    fscanf(fin,"%s\n",number);
    num=strlen(number);
    fclose(fin);
    begin=checkbegin[number[0]-50];
    stop=checkstop[number[0]-50];
    fscanf(fdict,"%s\n",print);
    while (stop>=print[0])
    {
        while (begin<=print[0] && stop >=print[0])
        {
            if (num==strlen(print))
            {
                printcheck=1;
                for (i=0;i<=num-1;i++)
                    if (check(i)==0)
                    {
                        printcheck=0;
                        break;
                    }
                if (printcheck==1)
                    {
                        fprintf(fout,"%s\n",print);
                        printcount++;
                    }
            }
            fscanf(fdict,"%s\n",print);
        }
    fscanf(fdict,"%s\n",print);
    }
    if (printcount==0)
        fprintf(fout,"NONE\n");
    fclose(fout);
    fclose(fdict);
    return(0);
}

laysent 2009-11-01 22:01 发表评论

USACO-Greedy Gift Givers

laysent — Sat, 31 Oct 2009 02:16:00 GMT

#include
#include
int main()
{
int i,j,ng,np,present,togive,gp,rp;
char names[10][15],tname[15];
int given[10],received[10];
    FILE *fin=fopen("gift1.in","r");
    FILE *fout=fopen("gift1.out","w");
fscanf(fin,"%d\n",&np);
for(i=0;i{
   fscanf(fin,"%s\n",names[i]);
   given[i]=received[i]=0;
}
for(i=0;i{
   fscanf(fin,"%s\n%d %d",tname,&present,&ng);
   for(gp=0;gp if(ng&&strcmp(names[gp],tname)==0) { given[gp]+=present-present%ng; togive=present/ng; break;}
   for(j=0;j   {
fscanf(fin,"%s\n",tname);
for(rp=0;rp if(strcmp(names[rp],tname)==0)
{
received[rp]+=togive;
break;
}
   }
}
for(i=0;i fprintf(fout,"%s %d\n",names[i],received[i]-given[i]);
return 0;
}

laysent 2009-10-31 10:16 发表评论

USACO-Transformations

laysent — Sat, 31 Oct 2009 02:15:00 GMT

#include "stdio.h"
#include "math.h"
FILE *fin,*fout;
int outputn=7,num,combinate=0;
long picture[9]={0},reflect[9]={0};

scan()
{
char propic[10];int i,k;
for (i=0;i<=(num-1);i++)
    {fscanf(fin,"%s\n",propic);
    for (k=0;k<=(num-1);k++)
        {if (propic[k]=='@')
        {picture[i]<<=1;picture[i]=picture[i]+1;}
        if (propic[k]=='-')
        picture[i]<<=1;}}
for (i=0;i<=(num-1);i++)
    {fscanf(fin,"%s\n",propic);
    for (k=0;k<=(num-1);k++)
        {if (propic[k]=='@')
        {reflect[i]<<=1;reflect[i]=reflect[i]+1;}
        if (propic[k]=='-')
        reflect[i]<<=1;}}
fclose(fin);
}

reflection()
{int i,j,count=0,correct=1;long propic,proref;
    while(count<=num-1)
    {
    for (i=num-1,j=0;i>=0;i--,j++)
    {
        propic=picture[count]>>i;
        proref=reflect[count]>>j;
        propic=propic&1;
        proref=proref&1;
        if (propic!=proref)
        { correct=0;break;}
            }
    count++;
    }
    if (correct==1 && outputn>4) outputn=4;
}

nochange()
{int i,correct=1;
for (i=0;i<=(num-1);i++)
    if (picture[i]!=reflect[i]) correct=0;
if (correct==1 && outputn>6) outputn=6;
}

rotation90()
{long propic[9],procheck=0;int i,correct=1,count=1;
while(count<=num)
{ for(i=0;i<=(num-1);i++)
    propic[i]=picture[i]; /* 复制原数*/
for (i=0;i<=(num-1);i++)
    {
    if (num-count-i>=0)
    propic[i]=propic[i]>>(num-count-i);
    else
    propic[i]=propic[i]<<(count+i-num);
    propic[i]=propic[i]&((int)(pow(2,i)));}    /* 将每个数第n位的值左移*/
procheck=0;
for (i=0;i<=(num-1);i++)
    procheck=propic[i]|procheck;      /* 将每个数第n位移好的值进行合并，成为旋转后的数*/
if(procheck!=reflect[count-1])
    {correct=0;break;}
count++;}
if (correct==1 && combinate==0) outputn=1;
if (correct==1&& combinate==1)outputn=5;
}

rotation180()
{int i,count=0,correct=1;long procheck=0,propic;
    while(count<=num-1)
    {
    procheck=0;
    for (i=0;i<=num-1;i++)
    {
        propic=picture[count]>>i;
        propic=propic&1;
        if (propic==1)
        {procheck<<=1;procheck=procheck+1;}
        else
        procheck<<=1;
    }
    if (procheck!=reflect[num-count-1])
    { correct=0;break;}
    count++;
    }
    if (correct==1 && outputn>2 && combinate==0) outputn=2;
    if (correct==1 && combinate==1) outputn=5;
}

rotation270()
{long propic[9],procheck=0;int i,j,correct=1,count=1;
while(count<=num)
{ for(i=0;i<=(num-1);i++)
    propic[i]=picture[i];
for (i=0,j=(num-1);i<=(num-1);i++,j--)
    {
    propic[i]=propic[i]>>(num-count-j);
    propic[i]=propic[i]&((int)(pow(2,j)));}
procheck=0;
for (i=0;i<=(num-1);i++)
    procheck=propic[i]|procheck;
if(procheck!=reflect[num-count])
    {correct=0;break;}
count++;}
if (correct==1 && outputn>3 && combinate==0) outputn=3;
if (correct==1&& combinate==1)outputn=5;
}

combination()
{int i,count=0;long procheck=0,propic;
    while(count<=num-1)
    {
    procheck=0;
    for (i=0;i<=num-1;i++)
    {
        propic=picture[count]>>i;
        propic=propic&1;
        if (propic==1)
        {procheck<<=1;procheck=procheck+1;}
        else
        procheck<<=1;
    }
    picture[count]=procheck;
    count++;
    }
    combinate=1;
    rotation90();
    rotation180();
    rotation270();
}

print()
{
    fout=fopen("transform.out","w");
    fprintf(fout,"%d\n",outputn);
}

main()
{int l;
fin=fopen("transform.in","r");
fout=fopen("transform.out","w");
fscanf(fin,"%d\n",&num);
scan();
rotation90();
if (outputn==7)
rotation180();
if (outputn==7)
rotation270();
if (outputn==7)
reflection();
if (outputn==7)
nochange();
if (outputn==7)
combination();
print();
return(0);
}

laysent 2009-10-31 10:15 发表评论

USACO-Friday the Thirteenth

laysent — Sat, 31 Oct 2009 02:14:00 GMT

#include "stdio.h"

int n;

int plus(wochen,tag,plusday)
int wochen[7],tag,plusday;
{
    tag+=plusday;
    if (tag>7) tag-=7;
    wochen[tag]+=1;
    return tag;
}

int leap(int y)
{
    int yon;
    yon=0;
    if (y%4==0)
    {if (y%100==0)
        {if(y%400==0)
            yon=1;
        else yon= 0;}
    else
        yon=1;}
    else
        yon=0;
    return (yon);
}

main()
{
    int i,month,year,day,week[7];
    FILE *fin;
    FILE *fout = fopen ("friday.out", "w");
    for (i=1;i<=7;i++)
        {week[i]=0;}
    fin=fopen("friday.in","r");
    fscanf(fin,"%d\n",&n);
    year=0;
    week[1]+=1;
    day=1;
    week[7]=0;
    month=1;
    while (year<=(n-1))
    {
        while (month<=12)
        {
            if (month==2)
                {
                if (leap(year+1900)==1)
                    day=plus(week,day,1);
                else
                    {day=plus(week,day,0);}
                }
            else
                {if (month==4 || month==6 || month==9 || month==11)
                    {
                    day=plus(week,day,2);
                    }
                else
                    {
                        day=plus(week,day,3);
                    }
                }
            month+=1;
        }
        month=1;
        year+=1;
    }
    week[day]-=1;
    for (i=1;i<=6;i++)
    fprintf(fout,"%d ",week[i]);
    fprintf(fout,"%d\n",week[7]);
exit(0);
}

laysent 2009-10-31 10:14 发表评论

USACO-Milking Cows

laysent — Sat, 31 Oct 2009 02:14:00 GMT

#include "stdio.h"
#define MAX 10000
int start[MAX],stop[MAX],pro;
int n;
FILE *fin,*fout;

main()
{struct new
    {int start,stop;}newline[MAX];
int i,j,count=0,arraymax=0,gapmax=0;
fin=fopen("milk2.in","r");
fout=fopen("milk2.out","w");
fscanf(fin,"%d\n",&n);
for (i=0;i    {
        fscanf(fin,"%ld %ld\n",&start[i],&stop[i]);
    }
for (i=0;i    {for(j=i+1;j        {
            if (start[i]>start[j])
            {
                pro=start[i];
                start[i]=start[j];
                start[j]=pro;
                pro=stop[i];
                stop[i]=stop[j];
                stop[j]=pro;
            }
        }
    }
newline[0].start=start[0];newline[0].stop=stop[0];
for (i=0;i    {
        if (start[i+1]<=newline[count].stop)
        {
            if (stop[i+1]>newline[count].stop)
                {newline[count].stop=stop[i+1];}
        }
        else
        {
            count++;
            newline[count].start=start[i+1];
            newline[count].stop=stop[i+1];
        }
    }
for (i=0;i    {
        if (arraymax<(newline[i].stop-newline[i].start)) arraymax=newline[i].stop-newline[i].start;
        if (gapmax<(newline[i+1].start-newline[i].stop)) gapmax=newline[i+1].start-newline[i].stop;
    }
if (arraymax<(newline[i].stop-newline[i].start)) arraymax=newline[i].stop-newline[i].start;
fprintf(fout,"%d %d\n",arraymax,gapmax);
return(0);
}

laysent 2009-10-31 10:14 发表评论

USACO-Broken Necklace

laysent — Sat, 31 Oct 2009 02:13:00 GMT

#include "stdio.h"
main()
{
    int promax,max,n,cut,count,turnover;
    char color[350],next,head;
    FILE *fin=fopen("beads.in","r");
    FILE *fout=fopen ("beads.out","w");
    fscanf(fin,"%d\n",&n);
    fscanf(fin,"%s\n",color);
    count=0; max=0;
    head=color[count];
    promax=1; cut=0;
    turnover=0;
    while (turnover>=0 && promax!=n)
    {
        if (count        if (head=='w' && color[count]!='w') head=color[count];
        if (head==color[count] ||color[count]=='w') promax+=1;
        else
        {if (cut==0) {cut=count;head=color[count];promax+=1;}
        else    {if (cut>0 && max                {max=promax; promax=1;
                count=cut; head=color[count];
                    while (color[count-1]=='w')
                    {count--;promax++;}
                count=cut; cut=0;
                    if (turnover==1) turnover=-1;
                }
            else
                {promax=1;
                count=cut;head=color[count];
                while (color[count-1]=='w')
                    {count--;promax++;}
                count=cut; cut=0;
                    if (turnover==1) turnover=-1;
                }
                }
        }
    }
    if (promax==n) fprintf(fout,"%d\n",promax);
    else    fprintf(fout,"%d\n",max);
    exit(0);
}

laysent 2009-10-31 10:13 发表评论

USACO-Your Ride Is Here

laysent — Sat, 31 Oct 2009 02:12:00 GMT

#include
main()
{
    FILE *fin = fopen ("ride.in", "r");
    FILE *fout = fopen ("ride.out", "w");
    char product[7], comet[7];
    int k,i,j,promod,commod;
    long pro,com;
    fscanf (fin, "%s\n", product);
    fscanf(fin,"%s\n",comet);
    k=strlen(product);
    pro=1;
    com=1;
    for (i=0;i<=k;i++)
        pro=pro*(product[i]-64);
    j=strlen(comet);
    for (i=0;i<=j;i++)
        com*=(comet[i]-64);
    promod=pro % 47;
    commod= com % 47;
    if (promod == commod) fprintf(fout, "GO\n");
    else fprintf(fout,"STAY\n");
    exit (0);

    }

laysent 2009-10-31 10:12 发表评论