Winter-1-D Max Sum 解题报告及测试数据

Time Limit:1000MS     

Memory Limit:32768KB

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 

以下是代码：

# include  <iostream>
# include  <cstring>
# include  <cstdio>
# include  <cstdlib>
using  namespace  std;
int  main(){
    int  n,tn;
    int  news,s,e,t;
    int  maxn,sum;
    cin>> n;
    for ( int  i=1;i<=n;i++){
        sum =  0 ;
        maxn = - 20000 ;
        news = s = e =  1 ;
        scanf( "%d" ,&tn);
        for ( int  j=1;j<=tn;j++){
            scanf( "%d" ,&t);
            sum+= t ;
            if (sum > maxn){  //目前的和大于记录和，更新和及起点终点
                maxn = sum;
                e = j;
                s = news;
            }
            if (sum <  0 ){ //目前的和小于0，则将起点定为下一个数，和为0
                sum= 0 ;
                news=j+ 1 ;
            }
        }
        printf( "Case %d:\n%d %d %d\n" ,i,maxn,s,e);
        if (i!=n)printf( "\n" );
    }
}