Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 157635    Accepted Submission(s): 36871

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

 

Author

Ignatius.L

 

 要注意最大值值可能是负数 ， so 不要用 result = 0 去比较取最大值 ；

看看能不能过 2 -7 7 这组。

（一开始我写的代码只能保证 在首项  非负是成立 ，so WA ）

#include<stdio.h>
#include<string.h>
int N ;
int a[100001] ;
int dp[100001] ;

int main()
{
    freopen ( "a.txt" ,"r" , stdin ) ;
    int T;
    int i , j ;
    int cnt = 0 ;
    int start , end ;
    int result ;
    scanf ( "%d" , &T ) ;
    while ( T-- )
    {
        scanf ( "%d" , &N );
        memset ( dp , 0 , sizeof(dp) );
        for ( i = 1 ; i <= N ; i++ )
            scanf ( "%d" , &a[i] ) ;

        for ( i = 1 ; i <= N ; i++ )
        {
                if ( dp[i - 1] + a[i] >= 0 && dp[i - 1] >= 0)
                    dp[i] = dp[i - 1] + a[i] ;
                else
                {
                    dp[i] = a[i] ;
                }
        }
        result = 1 ;
        for ( i = 2 ; i <= N ; i++ )
        {
            if( dp[result] < dp[i] )
                result = i ;
        }
        start = dp[result] ;
        for ( j = result ; j >= 1 ; j-- )
        {
            start-= a[j] ;
            if ( !start )
                end = j;
        }
        printf ( "Case %d:
%d %d %d
" , cnt + 1 , dp[result] , end , result ) ;
        cnt++ ;
        if ( T )
            printf ( "
" ) ;
    }
    return 0;
}