C++博客-purplest-随笔分类-others

poj 1114 Chemical Reactions

purplest — Tue, 04 Oct 2011 11:50:00 GMT

很烦的一道题，比赛时写了我3个小时，最后TLE。。。

下午才查出来了好几个bug。。。虽然题很烦，但确实磨练coding能力啊。。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct P
{
    char s[5];
    int n;
}node[11000], get[11000];

int cmp(const void *a , const void *b)
{
    return strcmp( (*(P *)a).s, (*(P *)b).s );
}

char str[11000], str1[11000];
int n, num;

void element(char *str, int s, int t, int k, P *p);

void insert(char *s, int k, P *p)
{
    int i;
    for ( i = 0 ; i < num ; i++ )
    {
        if ( !strcmp(s, p[i].s) )
            break;
    }
    if ( i < num )
        p[i].n+=k;
    else p[num].n+=k, strcpy(p[num++].s, s);
}

void sequence(char *str, int s, int t, int k, P *p)
{
    int i = s;
    while ( i < t )
    {
        if ( '(' ==str[i] )
        {
            int ti, tk;
            for ( ti = i+1, tk= 1 ; tk ; ti++ )
            {
                if ( '(' == str[ti] )
                    tk++;
                else if ( ')' == str[ti] )
                    tk--;
            }
            if ( '0' <= str[ti] && str[ti] <= '9' )
            {
                int ttk, te= ti;
                for  ( ttk = 0; '0' <= str[ti] && str[ti] <= '9' ; ti++ )
                    ttk= ttk*10+str[ti]-'0';
                element(str, i, te, k*ttk, p);
            }
            else element(str, i, ti, k, p);
            i= ti;
        }
        else if ( 'A' <= str[i] && str[i] <= 'Z' )
        {
            int tk= i;
            i++;
            if ( 'a' <= str[i] && str[i] <= 'z' )
                i++;
            if ( '0' <= str[i] && str[i] <= '9' )
            {
                int ttk;
                for  ( ttk = 0; '0' <= str[i] && str[i] <= '9' ; i++ )
                    ttk= ttk*10+str[i]-'0';
                element(str, tk, i, k*ttk, p);
            }
            else element(str, tk, i, k, p);
        }
    }
}

void element(char *str, int s, int t, int k, P *p)
{
    if ( '(' == str[s] && ')' == str[t-1] )
        sequence(str, s+1, t-1, k, p);
    else
    {
        char tmp[5];
        tmp[0] = str[s];
        if ( 'a' <= str[s+1] && str[s+1] <= 'z' )
            tmp[1]= str[s+1], tmp[2]= 0;
        else tmp[1]= 0;
        insert(tmp, k, p);
    }
}

void slove(char *str, P *p)
{
    int len = strlen(str), i= 0, j, k;
    num= 0;
    while ( i < len )
    {
        char ch;
        ch= str[i];
        if ( '0' <= ch && ch <= '9' )
        {
            for  ( k = 0; '0' <= str[i] && str[i] <= '9' ; i++ )
                k= k*10+str[i]-'0';
            for ( j = i ; str[j] && '+' != str[j] ; j++ );
            sequence(str, i, j, k, p);
            i= j+1;
        }
        else
        {
            for ( j = i+1 ; str[j] && '+' != str[j] ; j++ );
            sequence(str, i, j, 1, p);
            i= j+1;
        }
    }
}

int main()
{
    while ( EOF!= scanf("%s", str) )
    {
        memset(node, 0, sizeof(node));
        slove(str, node);
        int m= num;
        qsort(node, m, sizeof(P), cmp);
        scanf("%d", &n);
        while ( n-- )
        {
            scanf("%s", str1);
            memset(get, 0, sizeof(get));
            slove(str1, get);
            if ( m == num )
            {
                qsort(get, m, sizeof(P), cmp);
                int i;
                for ( i = 0 ; i <= num && node[i].n == get[i].n && !strcmp(node[i].s, get[i].s) ; i++ );
                if ( i > num )
                {
                    printf("%s==%s\n", str, str1);
                    continue;
                }
            }
            printf("%s!=%s\n", str, str1);
        }
    }
    return 0;
}

purplest 2011-10-04 19:50 发表评论

poj 1064 Cable master

purplest — Tue, 04 Oct 2011 11:47:00 GMT

很水的二分找答案，不过，需要注意精度的控制

#include <stdio.h>

const double eps= 1e-8;

int data[10010], m, n, ans;

bool judge(int mod)
{
    int cnt= 0;
    for ( int i = 0 ; i < n ; i++ )
        cnt+= data[i]/mod;
    return  cnt >= m;
}

int main()
{
    while ( 2 == scanf("%d%d", &n, &m ) )
    {
        int i, sum= 0;
        bool flag= true;
        double t;
        for ( i = 0 ; i < n ; i++ )
        {
            scanf("%lf", &t);
            data[i]= (int)((t+eps)*1000);
            sum+=data[i];
            (flag && sum < m*10) ? flag= true : flag= false;
        }
        if ( flag )
        {
            printf("0.00\n");
            continue;
        }
        int left= 0, right= 1000000000, mid;
        while ( left < right )
        {
            mid= left+right>>1;
            if ( judge(mid) )
                left= mid+1;
            else ans = right = mid-1;
        }
        int a= ans/1000, b= (ans-a*1000)/10, c= (ans-a*1000)%10;
        printf("%d.", a);
        if ( c > 9 )
            b++;
        if ( b < 10 )
            printf("0%d\n", b);
        else printf("%d\n", b);
    }
    return 0;
}

purplest 2011-10-04 19:47 发表评论

poj 1363 Rails

purplest — Mon, 03 Oct 2011 12:15:00 GMT

用栈模拟过，不过不是朴素的模拟，这方法在lrj的白书上看到的，比我们组队时想到的方法好的多了，而且才30+行。。。

#include <stdio.h>

const int N= 1010;

int n, stack[N], data[N];

int main()
{
    while ( scanf("%d", &n), n )
    {
        while ( scanf("%d", data), data[0] )
        {
            int i, m= 1, k= 0, top= 0;
            for ( i = 1 ; i < n ; i++ )
                scanf("%d", data+i);
            bool flag= true;
            while ( k < n )
            {
                if ( m == data[k] )
                {
                    m++;
                    k++;
                }
                else if ( top && stack[top] == data[k] )
                {
                    top--;
                    k++;
                }
                else if ( m < n )
                    stack[++top] = m++;
                else
                {
                    flag= false;
                    break;
                }
            }
            printf("%s\n", flag?"Yes":"No");
        }
        putchar(10);
    }
    return 0;
}

purplest 2011-10-03 20:15 发表评论

poj 1781 In Danger

purplest — Tue, 23 Aug 2011 16:44:00 GMT

约瑟夫变种，直接套约瑟夫的话会TLE

#include <stdio.h>

int num, n;
int work(int n)
{
    int t= 1073741824;
    while ( n < t )
        t>>=1;
    return (n-t)*2+1;
}

int main()
{
    while ( scanf("%de%d", &num, &n), num || n )
    {
        while ( n-- )
        {
            num*=10;
        }
        printf("%d\n", work(num));
    }
    return 0;
}

purplest 2011-08-24 00:44 发表评论

poj 1053 Set Me

purplest — Tue, 16 Aug 2011 16:46:00 GMT

水题，统计下重不重复即可

#include <stdio.h>
#include <string.h>

char str[20][10];
char t[50];

int work(int i, int j, int k, int m)
{
    int p = 0;
    memset(t, 0, sizeof(t));
    if (!t[ str[i][m] -'0' ])
    {
        p++;
        t[ str[i][m] -'0' ] = 1;
    }
    if (!t[ str[j][m] -'0' ])
    {
        p++;
        t[ str[j][m] -'0' ] = 1;
    }
    if (!t[ str[k][m] - '0' ])
    {
        p++;
        t[ str[k][m] -'0' ] = 1;
    }
    return p;
}

int main()
{
    while ( EOF != scanf("%s", str[0]) )
    {
        int i, j, k, l, count= 0;
        for ( i = 1 ; i < 12 ; i++ )
            scanf("%s", str[i]);
        printf("CARDS: ");
        for ( i = 0 ; i < 12 ; i++ )
            printf(" %s", str[i]);
        for ( i = 0 ; i < 12 ; i++ )
            for ( j = i+1 ; j < 12 ; j++ )
                for  ( k = j+1 ; k < 12 ; k++ )
                {
                    int p[4];
                    for ( l = 0 ; l < 4 ; l++ )
                    {
                        p[l]= work(i, j ,k, l);
                    }
                    if ( ( p[0] == 1 || p[0] == 3 ) && ( p[1] == 1 || p[1] == 3 ) && ( p[2] == 1 || p[2] == 3 ) && ( p[3] == 1 || p[3] == 3 ) )
                    {
                        count++;
                        if ( 1 == count )
                            printf("\nSETS:   1.  %s %s %s\n", str[i], str[j], str[k]);
                        else printf("        %d.  %s %s %s\n", count, str[i], str[j], str[k]);
                    }
                }
        if (!count) printf("\nSETS:   *** None Found ***\n");
        putchar(10);
    }
    return 0;
}

purplest 2011-08-17 00:46 发表评论

poj 2276 Model Rocket Height

purplest — Sun, 14 Aug 2011 16:20:00 GMT

公式题

#include <stdio.h>
#include <math.h>

const double pi=3.141592653;
double d, H, e=0.00000001;

int main()
{
    scanf("%lf%lf", &d, &H);
    double a, b, c, h;
    while ( scanf("%lf%lf%lf", &a, &b, &c), a > e && b > e && c > e )
    {
        a=tan(a/180*pi);
        b=tan(b/180*pi);
        c=tan(c/180*pi);
        h=H+d*sqrt(2.0/(1/a/a+1/c/c-2/b/b))+0.49999999;
        int t=h;
        printf("%d\n", t);
    }
    return 0;
}

purplest 2011-08-15 00:20 发表评论

poj 1050 To the Max

purplest — Fri, 12 Aug 2011 01:51:00 GMT

果的最大子矩阵问题，用到了最大子区间

#include <stdio.h>
#include <string.h>

int num[200][200];
int sub[200][200];
int n;

int maxnum(int *p)
{
    int max=0, sum=0;
    for ( int i = 1; i < n; i++ )
    {
        sum+=p[i];
        if ( sum < 0) sum= 0;
        else if ( sum > max) max= sum;
    }
    return max;
}

int max()
{
    int t[200];
    int i, j, k, m=0, l;
    for ( i = 1; i <= n; i++ )
        for ( j = i; j <= n; j++ )
        {
            for ( k = 1; k <= n; k++ )
                t[k]=sub[j][k]+sub[i-1][k-1]-sub[j][k-1]-sub[i-1][k];
            l=maxnum(t);
            if (l > m) m=l;
        }
    return m;
}

int main()
{
    while ( EOF != scanf("%d", &n) )
    {
        int i, j;
        for ( i = 1; i <= n ; i++ )
            for ( j = 1 ; j <= n ; j++ )
                scanf("%d", num[i]+j);
        memset(sub, 0, sizeof(sub));
        sub[1][1]=num[1][1];
        for ( i = 1; i <= n ; i++ )
            for ( j = 1; j <= n ; j++ )
                sub[i][j]=sub[i-1][j]+sub[i][j-1]+num[i][j]-sub[i-1][j-1];
        printf("%d\n", max());
    }
    return 0;
}

purplest 2011-08-12 09:51 发表评论

poj 2850 Stacking Cylinders

purplest — Wed, 10 Aug 2011 16:43:00 GMT

读懂题后其实就是一道数学题了
hit:The distance between adjacent centers will be at least 2.0 (so the cylinders do not overlap) and at most 3.4...
3.4=1+1+1.4，1.4=sqrt(2);

#include <stdio.h>
#include <math.h>

double num[20][20], numy[20][20];

int n;

void getpoint( double x1, double y1, double x2, double y2 , double & x, double & y )
{
    double a, b, c, d, e, f;
    a= x2-x1;
    b= y2-y1;
    c= (x1+x2)/2.0;
    d= (y1+y2)/2.0;
    e= (y1-y2)/2.0;
    f= (x1-x2)/2.0;
    y=sqrt( (4-e*e-f*f)*a*a/(a*a+b*b) ) + d;
    x= ( b*( d-y )/a )+c;
}

int main()
{
    scanf("%d", &n);
    for ( int l= 1; l <= n; l++ )
    {
        int t;
        scanf("%d", &t);
        int i, j;
        for ( i = 0; i < t ; i++ )
        {
            scanf("%lf", num[t-1]+i);
            numy[t-1][i]=1.0;
        }
        for ( j = t-1; j > 0 ; j-- )
        {
            for ( i = 0; i < t; i++ )
            {
                getpoint(num[j][i], numy[j][i], num[j][i+1], numy[j][i+1], num[j-1][i], numy[j-1][i]);
            }
        }
        printf("%d: %.4f %.4f\n", l, num[0][0], numy[0][0]);
    }
    return 0;
}

purplest 2011-08-11 00:43 发表评论

poj 2853 Sequence Sum Possibilities

purplest — Wed, 10 Aug 2011 16:39:00 GMT

就是等差数列，模拟会TLE的

#include <stdio.h>
#include <math.h>

int main()
{
    int cases=0;
    while (scanf("%d", &cases) != EOF){
        while (cases--){
            int k;
            int n, i, count=0;
            scanf("%d%d",&k, &n);
            int num= (int)sqrt(2*n+0.1)+1;
            for ( i = 1; i < num; i++ )
            {
                if( (2*n%i)==0 && (2*n/i+1-i) > 0 &&  (2*n/i+1-i)%2 == 0)
                count++;
            }
            printf("%d %d\n", k, count-1);
        }
    }
    return 0;
}

purplest 2011-08-11 00:39 发表评论

poj 1046 Color Me Less

purplest — Wed, 10 Aug 2011 16:32:00 GMT

水题

#include <stdio.h>

int p[20][5], data[5];

int dis(int i)
{
    return (p[i][0]-data[0])*(p[i][0]-data[0])+(p[i][1]-data[1])*(p[i][1]-data[1])+(p[i][2]-data[2])*(p[i][2]-data[2]);
}

int mindist()
{
    int i, min=10000000, dist, k;
    for ( i = 0; i < 16; i++ )
    {
        dist=dis(i);
        if ( min > dist )
        {
            min= dist;
            k= i;
        }
    }
    return k;
}

int main()
{
    int i;
    for ( i = 0; i < 16; i++ )
    {
        scanf("%d%d%d", p[i], p[i]+1, p[i]+2);
    }
    while ( scanf("%d%d%d", data, data+1, data+2), data[0]!= -1 && data[1] != -1 && data[2] != -1 )
    {
        int  j= mindist();
        printf("(%d,%d,%d) maps to (%d,%d,%d)\n", data[0], data[1], data[2], p[j][0], p[j][1], p[j][2]);
    }
    return 0;
}

purplest 2011-08-11 00:32 发表评论

poj 1051 P,MTHBGWB

purplest — Wed, 10 Aug 2011 16:28:00 GMT

一道摩尔电码翻译的题，用的模拟，还好数据量不大，如果数据量大了，得上map，或者字典树了。。

#include <cstdio>
#include <iostream>
#include <string>
using namespace std;

int n;
string chara[40]={"12", "2111", "2121", "211", "1", "1121", "221", "1111", "11", "1222", "212", "1211", "22", "21", "222", "1221",
"2212", "121", "111", "2", "112", "1112", "122", "2112", "2122", "2211", /*符号*/ "1122"/*_*/, "2221",/*.*/ "1212",/*,*/ "2222"/*?*/};

int main()
{
    while ( EOF != scanf("%d", &n) )
    {
        int q;
        for (q=1; q<=n; q++)
        {
            int num[100];
            char str[510];
            string ch="";
            scanf("%s", str);
            int i, j, k= 0, p= 0, len;
            for ( i = 0; str[i] != 0; i++ )
            {
                if ( 'A' <= str[i] && 'Z' >= str[i] )
                {
                    num[i]= chara[str[i]-'A'].length();
                    ch+=chara[str[i]-'A'];
                }
                else if ( str[i] == '_' )
                {
                    num[i]= chara[26].length();
                    ch+=chara[26];
                }
                else if ( str[i] == '.' )
                {
                    num[i]= chara[27].length();
                    ch+=chara[27];
                }
                else if ( str[i] == ',' )
                {
                    num[i]= chara[28].length();
                    ch+=chara[28];
                }
                else if ( str[i] == '?' )
                {
                    num[i]= chara[29].length();
                    ch+=chara[29];
                }
                k++;
            }
            printf("%d: ", q);
            for ( i= k-1; i >= 0; i--)
            {
                len= num[i];
                string t;
                for ( j= 0; j < len; j++)
                    t+=ch[j+p];
                for ( j = 0; j < 30 && t != chara[j]; j++);
                if ( 0 <= j && 26 > j) printf("%c", j+'A');
                else if ( 26 == j ) printf("_");
                else if ( 27 == j ) printf(".");
                else if ( 28 == j ) printf(",");
                else if ( 29 == j ) printf("?");
                p+=len;
            }
            putchar(10);
        }
    }
    return 0;
}

purplest 2011-08-11 00:28 发表评论