HDOJ-----1003Max Sum

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 217002 Accepted Submission(s): 51163

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

这个和1231最大子序列那道题基本一样的，就输出改了一下格式，参考上篇博客，就不重复了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 100010using namespace std;int s[maxn];int main(){int T, t, n, m, v, st, en, kcase = 1;cin >> T;while(T--){scanf("%d", &n);memset(s, 0, sizeof(s));st = en = v = 1;for(int i = 1; i <= n; i++){scanf("%d", &s[i]);if(i == 1){t = m = s[1];continue;}if(t + s[i] < s[i]){t = s[i];v = i;}else{t += s[i];}if(t > m){st = v;en = i;m = t;}}printf("Case %d:\n%d %d %d\n", kcase++, m, st, en);if(T){printf("\n");}}return 0;}

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