Codeforces-33C. Wonderful Randomized Sum

传送门

N个数，允许将前连续任意个数变化为其相反数，也允许把后连续任意个数变为相反数，求最大和

令dp[i][0]前i个数操作后能得到的最大值，dp[i][1]出去前i-1个数操作后能得到的最大值

注意初始化，不然对于答案为无需操作的情况会出错

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

const int maxn = 1e5 + 10;
int A[maxn];
int dp[maxn][2];
int sum[maxn];

int N;

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        scanf("%d", &A[i]);
        sum[i] = sum[i - 1] + A[i];
    }
    int tmp = -INF;
    for (int i = 1; i <= N; i++) {
        tmp = max(tmp, -2 * sum[i]);
        dp[i][0] = sum[i] + tmp;
    }
    tmp = -INF;
    for (int i = N; i > 0; i--) {
        tmp = max(tmp, -2 * (sum[N] - sum[i - 1]));
        dp[i][1] = (sum[N] - sum[i - 1]) + tmp;
    }
    int ans = sum[N];//
    for (int i = 0; i <= N; i++) {
        ans = max(ans, dp[i][0] + dp[i + 1][1]);
    }
    printf("%d\n", ans);
    return 0;
}

原文地址：https://www.cnblogs.com/xFANx/p/8448187.html