C. Four Segments 前缀后缀

C. Four Segments

这种分成了三个节点一般都可以处理一下前缀处理一下后缀，或者处理一下前面的这个点，处理一下后面的这个点，然后再枚举中间这个点。

如果和中间这个点有关的，那么就可以换一下顺序，先枚举中间这个点，然后处理前面和后面的点。

这个是先枚举中间这个点，然后往前面贪心，往后面贪心。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#include <list>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 5e3 + 10;
ll sum[maxn], a[maxn];

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), sum[i] = sum[i - 1] + a[i];
    ll ans = 0, x = 1, y = 1, z = 1;
    for(int i=1;i<=n;i++)
    {
        ll res1 = 0, mark1 = 1;
        for (int j = i; j <= n + 1; j++) {
            ll num = sum[j - 1] * 2 - sum[i - 1] - sum[n];
            if (num > res1)    res1 = num, mark1 = j;
        }
        ll res2 = 0, mark2 = 1;
        for (int j = 1; j <= i; j++) {
            ll num = 2 * sum[j - 1] - sum[i - 1];
            if (num > res2) res2 = num, mark2 = j;
        }
        if(res1+res2>ans)
        {
            ans = res1 + res2;
            x = mark2, y = i, z = mark1;
            // printf("ans=%d\n", ans);
        }
    }
    printf("%lld %lld %lld\n", x - 1, y - 1, z - 1);
    return 0;
}