ZOJ-3023

来源：互联网发布：飞月捏脸数据编辑：程序博客网时间：2024/05/08 06:11

数组大小才100，直接硬算，全部枚举

#include<stdio.h>#include<stdlib.h>int cmp3023(const void *i1, const void *i2){    return *(int *) i1 - *(int *) i2;}int main(){    int n, m, i, j;    int a[100], b[100];    while (scanf("%d %d", &n, &m), n || m)    {        int sumt = 0, sumh = 0;        for (i = 0; i < n; i++)        {            scanf("%d", &a[i]);            sumt += a[i];        }        for (i = 0; i < m; i++)        {            scanf("%d", &b[i]);            sumh += b[i];        }        if ((sumt - sumh) % 2 != 0)        {            printf("-1\n");            continue;        }        qsort(a, n, sizeof(int), cmp3023);        qsort(b, m, sizeof(int), cmp3023);        int d = (sumt - sumh) / 2, found = 0;        for (i = 0; i < n && !found; i++)            for (j = 0; j < m; j++)                if (a[i] - b[j] == d)                {                    printf("%d %d\n", a[i], b[j]);                    found = 1;                    break;                }        if (!found)            printf("-1\n");    }    return 0;}

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