HDOJ 1003 Max Sum【MSS】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158875    Accepted Submission(s): 37166

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: 求最大字段和，并给出字段的起始点和终点。

题解:DP之，模板题——O(n)的算法。

AC代码:

#include<iostream>#define N 100000+5using namespace std;int dp[N],t,n;int main(){    cin.sync_with_stdio(false);    cin>>t;    for(int k=1;k<=t;k++){        cin>>n;        for(int i=0;i<n;i++)cin>>dp[i];        int sum=dp[0],tmp=dp[0],s=1,b=1,d=1;        for(int i=1;i<n;i++){            if(tmp>=0)                tmp+=dp[i];            else                tmp=dp[i],s=i+1;            if(tmp>sum) sum=tmp,d=i+1,b=s;        }        cout<<"Case "<<k<<":"<<endl;        cout<<sum<<" "<<b<<" "<<d<<endl;        if(k!=t)            cout<<endl;    }    return 0;}