事实上这连续发表的三篇是一模一样的思路，我就厚颜无耻的再发一篇吧！
题目链接：
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欢迎光临天资小屋：
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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
代码例如以下：
#include#define INF 0x3fffffff#define M 100000+17int a[M];int main(){ int n, i, T, k = 0; while(~scanf("%d",&T)) { while(T--) { scanf("%d",&n); int s = 1, e = 1, t = 1; int sum = 0, MAX = -INF; for(i = 1; i <= n; i++) { scanf("%d",&a[i]); sum+=a[i]; if(sum > MAX) { s = t; e = i; MAX = sum; } if(sum < 0) { t = i+1; sum = 0; } } printf("Case %d:\n",++k); printf("%d %d %d\n",MAX,s,e); if(T!=0) printf("\n"); } } return 0;}