USACO s1.1.2Greedy Gift Givers(字符串）

题意：计算每一位friend交换钱后，每一位手上拿着的钱是多少。

思路：二维数组，字符串。初始化。读懂题目就好了，对简单的题要快速码完，细心点，

/*
TASK:gift1
LANG:C++
ID:huibaochen
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
const int maxn = 15;
const int minn = 20;
using namespace std;

int main()
{
    freopen ("gift1.in", "r", stdin);
    freopen ("gift1.out", "w", stdout);
    int t, m, n, div, mod;
    char name[maxn][minn], name1[maxn][minn];
    int name2[maxn];
    while (scanf("%d", &t) != EOF){
        memset (name, 0, sizeof(name));
        memset (name2, 0, sizeof(name2));
        for (int i = 0; i < t; i++){
            scanf("%s", name[i]);
        }
        for (int i = 0; i < t; i++){
            memset (name1, 0, sizeof(name1));
            mod = div = 0;
            scanf ("%s", name1[0]);
            scanf ("%d%d", &m, &n);
            if ((m == 0) && (n == 0))
               continue;
            else{
                if(n != 0){
                div = m / n;
                mod = m % n;
                for (int j = 1; j <= n; j++)
                    scanf("%s", name1[j]);
                for (int j = 1; j <= n; j++){
                    for (int k = 0; k < t; k++){
                        if ((strcmp(name1[j], name[k])) == 0)
                            name2[k] = name2[k] + div;
                    }

                }
                 for (int j = 0; j < t; j++){
                    if((strcmp(name[j], name1[0])) == 0)
                    name2[j] = name2[j] - m + mod;
                }
                }
        }
        }
        for (int i = 0; i < t; i++){
            printf("%s %d\n", name[i], name2[i]);
        }
    }
    return 0;
}

 

不要在细节上卡题。