hdu 1003 Max Sum （最大连续子序列）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158643    Accepted Submission(s): 37111

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<iostream>#include<cstdio>#include<cstring>#define INF 0xfffffffusing namespace std;int a[100005];int dp[100005];int main(){    int t;    int n;    int max;    int s,ss,e;    scanf("%d",&t);    for(int k=1; k<=t; k++)    {        memset(dp,0,sizeof(dp));        scanf("%d",&n);        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        max=-INF;        s=e=ss=1;        for(int i=1; i<=n; i++)        {            dp[i]=dp[i-1]+a[i];            if(dp[i]<a[i])            {                dp[i]=a[i];                ss=i;            }            if(dp[i]>max)            {                max=dp[i];                s=ss;                e=i;            }        }        printf("Case %d:\n%d %d %d\n",k,max,s,e);        if(k<t)            printf("\n");    }    return 0;}