hdu 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 172541    Accepted Submission(s): 40197

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

 

Author

Ignatius.L

 

题意就是求最大子段和   状态转移方程dp[i]=max(dp[i-1]+ss[i],ss[i])

就是找初始位置被坑了几次  没有考虑全是负数的情况  

AC代码如下：

#include <stdio.h>
#include "string.h"
int main(void)
{
    int m,leap=1;
    scanf("%d",&m);
    while(m--)
    {
        int n,end,start;
        int ss[100001],max;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&ss[i]);
        max=ss[0];
        end=0;
        for(int i=1;i<n;i++)
        {
            if(ss[i-1]>0) ss[i]+=ss[i-1];
            if(ss[i]>max){
                max=ss[i];
                end=i;//找到最大子段的末位
            }
        }
        if(max<0) start=end; //如果最大子段和小于0,说明所有元素都是负的,最大子段就是那个最大的负数
        else{
            for(int i=end;i>=0;i--)
                if(ss[i]>=0) start=i;
                else break;
        }
        printf("Case %d:\n%d %d %d\n",leap++,max,start+1,end+1);
        if(m) printf("\n");
    }
}