HDU1003

PS: 这道题目很不错，虽然不难但是很经典，整理好思路再编码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x7ffffff;
const int maxn = 100010;
//Accepted	1003	15MS	1020K	1114 B	C++	Achiberx
int d[maxn];
int x[maxn];
int T, n;
void dp(int t)
{
    int idx;
    int st, endx;
    int maxv = -INF;
    memset(x, 0, sizeof(x));
    for(int i = 1; i <= n; i++) {
        if(i==1) x[i] = d[i];
        else {
            x[i] = max(d[i], x[i-1]+d[i]);
        }
        if(x[i] > maxv) {
            maxv = x[i];
            idx = i;
            endx = i;
        }
    }
    st = endx;
    for(int i = idx-1; i >= 1; i--) {
        if(x[i] >= 0)  {
            st = i;
        }
        else break;
    }
    printf("Case %d:\n", t);
    printf("%d %d %d\n", maxv, st, endx);
    if(t!=T) printf("\n");
}

int main()
{
    scanf("%d", &T);
    for(int t = 1; t <= T; t++) {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &d[i]);
        }
        dp(t);
    }
    return 0;
}
/*********
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
4 0 0 0 0
5 -1 -2 0 -3 -4
5 2 -1 3 4 0
7 6 -1 -7 5 4 -7 9
5 -3 -2 -1 0 -5
4 0 0 0 0
12 -1 2 5 -1 3 -10 -2 3 4 -1 3 -10
********/

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HDU1003

原文：http://blog.csdn.net/achiberx/article/details/23216283