HDU 1003 Max Sum

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

求最大子段和，基础题！

#include <stdio.h>
int main()
{
    int t;
    int n;
    int a[100001];
    scanf("%d", &t);
    for(int i=1; i<=t; i++){
        scanf("%d",&n);
        for(int j=1; j<=n; j++){
            scanf("%d",&a[j]);
        }
        int p1,p2;
        p1=p2=1;
        int pos=1;
        int temp=a[1];
        int res=a[1];
        for(int j=2; j<=n; j++){
            //如果之前是负的话，就重新开始存
            //-1 3 -1 2
            //第一次是-1的时候就重新存
            //第二次则不需要，因为之前的和为正数 
            if(a[j]+temp<a[j]){
                temp=a[j];
                pos=j; 
            }else{
                temp+=a[j];
            }
            //每次发现最大的时候则存放起来 
            if(temp>=res){
                res=temp;
                p1=pos;
                p2=j;
            }
        }
        if(i!=1)printf("
");
        printf("Case %d:
", i);
        printf("%d %d %d
", res, p1, p2);
    }
    return 0;
}