Codeforces-462A. A Compatible Pair

传送门

B从由两个数列中各挑出一个数相乘，他想让乘积最大化，A想让乘积最小化，他可以抹去一个数。求最终B得到的乘积

场上疯狂hack...

由于数据十分的小，我当时直接暴力求解A要抹去的数，然后再暴力求乘积即可。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int maxn = 60;
int N, M;

LL A[maxn], B[maxn];

int main() {
    scanf("%d%d", &N, &M);
    for (int i = 1; i <= N; i++) {
        scanf("%lld", &A[i]);
    }
    for (int i = 1; i <= M; i++) {
        scanf("%lld", &B[i]);
    }
    LL I = -1000000000000000001;
    LL tmp = I;
    int flag = 1;
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++) {
            LL t = A[i] * B[j];
            if (t > tmp) {
                tmp = t;
                flag = i;
            }
        }
    }
    tmp = I;
    for (int i = 1; i <= N; i++) {
        if (i == flag) continue;
        for (int j = 1; j <= M; j++) {
            LL t = A[i] * B[j];
            if (t > tmp) {
                tmp = t;
            }
        }
    }
    printf("%lld
", tmp);
    return 0;
}