HDU 1003 Max Sum（最大子列和）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

代码如下：

#include <cstdio>#define INF 0x3f3f3f3fint main(){    int t;    int n, tt;    scanf("%d",&t);    int k = 0;    while(t--)    {        scanf("%d",&n);        int sum = 0, maxx = -INF;        int start = 1, endd = 1, pos = 1;        for(int i = 1; i <= n; i++)        {            scanf("%d",&tt);            sum+=tt;            if(sum > maxx)            {                start = pos;                endd = i;                maxx = sum;            }            if(sum < 0)            {                sum = 0;                pos = i+1;            }        }        printf("Case %d:\n",++k);        printf("%d %d %d\n",maxx,start,endd);        if(t)            printf("\n");    }    return 0;}