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hdoj1003--Max Sum

时间:2014-03-01 03:20:16      阅读:433      评论:0      收藏:0      [点我收藏+]
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
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#include<iostream>
#include<cstdio>
using namespace std;

void solve()
{
    int T, i, max,sum,n,num,begin,end,t,con;
    while(cin>>T) {
        for(i = 1; i <= T; i++) {
            max = -1000000;
            sum = 0;
            n = con = 0;
            cin>>n;
            num = n;
            begin = end = 0;
            while(n--) {
                cin>>t;
                sum += t;
                con++;
                if(sum > max) {
                    begin = con;
                    max = sum;
                    end = num - n;
                }
                if(sum < 0) {
                    sum = 0;
                    con = 0;
                }
            }
            cout<<"Case "<<i<<":"<<endl<<max<<" "<<end-begin+1<<" "<<end<<endl;  
            if(i != T) cout<<endl;  
        }
    }
}

int main()
{     
    solve();
    return 0;
}
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hdoj1003--Max Sum

原文:http://www.cnblogs.com/xueda120/p/3573868.html

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