hdu1003(最大连续子列和 )

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 268751    Accepted Submission(s): 63871

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：动态规划。考虑数a[i]的决策，有两种情况：1.连续子列a[k]...a[i-1]的和加上a[i]，

2.a[i]为一个新子列的开端，

令dp[i]为以a[i]结尾的最大连续子列和，那么dp[i]=max(dp[i-1]+a[i],a[i])；

化简即if(dp[i-1]<0) dp[i]=a[i];else dp[i]=dp[i-1]+a[i];

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[200000];
int dp[200000];
int main()
{
    int T,N,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        if(cas) printf("\n");
        scanf("%d",&N);
        for(int i=0;i<N;i++) scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        dp[0]=a[0];
        int ans=a[0],l=0,r=0,cur=0;
        for(int i=1;i<N;i++)
        {
            if(dp[i-1]<0) {cur=i;dp[i]=a[i];}
            else dp[i]=dp[i-1]+a[i];
            if(ans<dp[i])
            {
                l=cur;
                ans=dp[i];
                r=i;
            }
        }
        printf("Case %d:\n%d %d %d\n",++cas,ans,l+1,r+1);
    }
    return 0;
}