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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/xFANx/p/8449061.html
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