Max Sum |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 191 Accepted Submission(s): 98 |
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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.
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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).
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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.
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Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
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Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
#include <iostream>
using namespace std; int main() { int T,N,max,sum,start,end,temp,t; cin>>T; int h = T; while(T--) { sum = 0; start = 1; end = 1; temp = 1; max = -1001; cin>>N; for(int i = 1; i <= N; i++) { cin>>t; sum +=t; if(sum>max) { max = sum; end = i; start = temp; } if(sum<0) { sum = 0; temp = i+1; } } if(T!=0) cout<<"Case "<<h-T<<":"<<endl<<max<<" " <<start<<" "<<end<<endl<<endl; else cout<<"Case "<<h-T<<":"<<endl<<max <<" "<<start<<" "<<end<<endl; } return 0; }
本文转自NewPanderKing51CTO博客，原文链接：http://www.cnblogs.com/newpanderking/archive/2011/08/03/2125824.html，如需转载请自行联系原作者
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