最大连续和2——dp

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 161115    Accepted Submission(s): 37730

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

#include<stdio.h>

int num[100001];
int book[100001];
int main()
{
    void res(int num[],int n);
    int T,cases=0;
    scanf("%d",&T);
    while(T--)
    {
        cases++;
        int total;
        scanf("%d",&total);
        int i;
        for(i=1;i<=total;i++)
            scanf("%d",&num[i]);
        if(cases!=1)
            printf("\n");
        printf("Case %d:\n",cases);
        res(num,total);
    }
    return 0;
}
void res(int num[],int n)
{


    int i;
    book[1]=1;
    for(i=2;i<=n;i++)
    {
        if(num[i]+num[i-1]>=num[i])
        {
            num[i]=num[i]+num[i-1];
            book[i]=book[i-1];
        }
        else
            book[i]=i;
    }
    int max=num[1],start=1,end=1;
    for(i=1;i<=n;i++)
    {
        if(max<num[i])
        {
            max=num[i];
            start=book[i];
            end=i;
        }
    }
    printf("%d %d %d\n",max,start,end);

}

dp思想，每一次都把当前的数值就行更新，相当于每一次都对这个数据进行记录。