UVALive6969 ZOJ3819 Average Score

Regionals 2014 >> Asia - Mudanjiang

问题链接:UVALive6969 ZOJ3819 Average Score。基础训练题,用C语言编写。

题意:Bob班里有n个人(包括Bob),邻班有n个人。若Bob不在本班,加入到邻班,两班的平均分都会增加。给出Bob班里n-1个人的分数以及邻班m个人的分数,问Bob分数可能的最小值和最大值是多少。

假设Bob分数为x,Bob班分数和为suma,邻班分数和为sumb,则必须满足suma/(n-1) >= x >= sumb/(m + 1)。Bob的成绩必须在整数域中进行考虑,比邻班平均成绩的要多1分,比本班的平均成绩要少1分。

AC通过的C语言程序如下:

/* UVALive6969 ZOJ3819 Average Score */

#include <stdio.h>

int main(void)
{
    int t, n, m, suma, sumb, val, min, max, i;

    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);

        suma = 0;
        for(i=1; i<=n-1; i++) {
            scanf("%d", &val);
            suma += val;
        }
        sumb = 0;
        for(i=1; i<=m; i++) {
            scanf("%d", &val);
            sumb += val;
        }

        min = sumb / m + 1;
        max = (suma % (n-1) == 0) ? suma / (n-1) - 1 : suma / (n-1);

        printf("%d %d\n", min, max);
    }

    return 0;
}
时间: 2024-10-11 10:36:51

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