简单dp hdu1003

Max Sum

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

 

这题蛮悲剧的 总共交了13次 可能理解的还是不够深吧 有几个地方一直没注意

#include<stdio.h>
int a[100001];
int main()
{
    long t,n,nowmax,premax,x,y,k = 0,i,start;
    scanf("%ld", &t);
    while(t--)
    {
        k++;
        scanf("%ld", &n);
        for(i = 1 ; i <= n ; i++)
            scanf("%ld", &a[i]);          
        premax = -999;//因为输入数据大于-1000，premax比-1000大就行了
        nowmax = 0;
        start = 1;
        x = 1;
        y = 1;
        for(i = 1 ; i <= n ; i++)
        {
            nowmax+=a[i];
            if(nowmax>premax)
            {
                y = i;
                start = x;//之前没有定义start 想着用下面的x就可以找出初始位置了 如果之前已经找到最大的premax 而后面又有负的nowmax 那x就不为初始位置 所以这里要跟着变动
                premax = nowmax;
            }           
            if(nowmax<0)//这里本来写的else if 如果nowmax是负的并且比premax大 就不对了
            {
                x = i+1;
                nowmax = 0;
            }
        }
        printf("Case %ld:\n",k);
        printf("%ld %ld %d\n", premax,start,y);
        if(t!=0)
            printf("\n");
    }
    return 0;
}