Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence.
If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

这里再给出一些测试数据：

4 0 0 2 0 —— 2 1 3
6 2 7 -9 5 4 3 —— 12 1 6
4 0 0 -1 0 —— 0 1 1
7 -1 -2 -3 -2 -5 -1 -2 —— -1 1 1
6 -1 -2 -3 1 2 3 —— 6 4 6

5 -3 -2 -1 -2 -3 —— -1 3 3

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<cmath>

using namespace std;
const int maxn = 100000 + 150;

int t;
int n;
int temp;
int start;
int last;
int a[maxn];
int ans;
int sum;

int main()
{
    scanf("%d", &t);
    for(int cas=1; cas<=t; cas++)
    {
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
            scanf("%d", &a[i]);
        sum = a[1];
        ans = a[1];//开始将其赋值为0 。。WA。。
        temp = 1;
        start = 1;
        last = 1;//以为last不用初始也可以。。WA。。
        for(int j=2; j<=n; j++)
        {
            if( sum<0 )
            {
                temp = j;
                sum = 0;
            }
            sum = sum + a[j];
            if( sum>ans )
            {
                ans = sum;
                last = j;
                start = temp;
            }
        }
        printf("Case %d:\n", cas);
        printf("%d %d %d\n", ans, start, last);
        if( cas<t )
            printf("\n");
    }

    return 0;
}

HDU 1003：Max Sum（DP）