Max Sum（最大子串和）

Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

Description

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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.

 

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Input

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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).

 

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Output

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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.

 

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Sample Input

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2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

 

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Sample Output

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Case 1:

14 1 4

 

Case 2:

7 1 6

 

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Analysis

 

　　状态转移方程 dp[i] = max( dp[i-1] + a[i] , a[i] )

 

 

#include <stdio.h>
#include <string.h>

int a[100002];

main()
{
    int T,N,kkk=1,i,j;

    scanf("%d",&T);
    while(T--)
    {
        int max,max_s,max_e,sum,sum_s;
        //memset
        printf("Case %d:\n",kkk++);

        scanf("%d",&N);
        for(i=1;i<=N;i++)
        {
            scanf("%d",&a[i]);
        }

        max = sum = a[1];
        max_s = max_e = sum_s = 1;

        //printf("sum = %d  max = %d\n",sum,max);
        for(i=2;i<=N;i++)
        {
            if(sum + a[i] < a[i])
            {
                sum = a[i] ;
                sum_s = i ;
            }
            else sum = sum + a[i] ;

            if(sum > max)
            {
                max = sum ;
                max_s = sum_s;
                max_e = i;
            }
            //printf("sum = %d  max = %d\n",sum,max);
        }
        printf("%d %d %d\n",max,max_s,max_e);
        if(T)
        printf("\n");
    }
}