HDU 1003 Max Sum （最大连续子序和）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 156637    Accepted Submission(s): 36628

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

 分析：这题是求最大的连续的子序列的和，输出最大连续子序和、起始下标、终止下标。与HDU的1231很类似。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
using namespace std;
const int MAX = 10010;
int dp[MAX],a;
int main()
{
    int T,i,j,n,left,right,maxn,flag,num;
    scanf("%d",&T);
    for(j=1;j<=T;j++)
    {
        scanf("%d",&n);
        maxn=-9999;
        num=0;
        left=right=flag=1;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a);
            num+=a;
            if(num>maxn)
            {
                maxn=num;
                left=flag;
                right=i;
            }
            if(num<0)
            {
                num=0;
                flag=i+1;
            }
        }
        if(j!=1) printf("\n");
        printf("Case %d:\n",j);
        printf("%d %d %d\n",maxn,left,right);
    }
    return 0;
}