HDOJ-----1003Max Sum

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

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 100010
using 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;
}