Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 256790    Accepted Submission(s): 61007

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

中文描述

给定序列a [1]，a [2]，a [3] ...... a [n]，您的任务是计算子序列的最大和。例如，给定（6，-1,5,4，-7），该序列中的最大和为6 +（-1）+ 5 + 4 = 14。

 

输入

输入的第一行包含一个整数T（1 <= T <= 20），这意味着测试用例数。然后T行跟随，每行以数字N（1 <= N <= 100000）开始，然后跟随N个整数（所有整数在-1000和1000之间）。

 

产量

对于每个测试用例，您应该输出两行。第一行是“Case＃：”，＃表示测试用例的编号。第二行包含三个整数，序列中的最大和，子序列的开始位置，子序列的结束位置。如果有多个结果，输出第一个结果。在两种情况之间输出空白行。

 

样品输入

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

样品输出

案例1：14 1 4 案例2：7 1 6
代码：
#include <stdio.h>int main(){int T,n,m;int first,end,x;int i,j,max,now;scanf("%d",&T);for(i=1;i<=T;i++){scanf("%d%d",&n,&m);max=now=m;first=x=1;for(j=2;j<=n;j++){scanf("%d",&m);if(now+m<m){now=m;x=j;}elsenow+=m;if(now>max){max=now;first=x;end=j;}}printf("Case %d:\n",i);printf("%d %d %d\n",max,first,end);if(i!=T)printf("\n");}return 0; }