杭电OJ 1003 Max Sum
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 240571 Accepted Submission(s): 56812
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>//#define LOCALusing namespace std;int a[100010];int main(int argc,char ** argv){ #ifdef LOCAL freopen("data.in","r",stdin); #endif int T,N,sum,start,temp,end,max; cin>>T; int count = 1; while(T--) { cin>>N; sum = 0; start = end = temp = 1; max = -10000; for(int i = 1; i <= N; i++) { cin>>a[i]; sum += a[i]; if(sum > max) { max = sum; start = temp; end = i; } if(sum < 0) { sum = 0; temp = i + 1; } } cout<<"Case "<<count++<<":"<<endl; cout<<max<<" "<<start<<" "<<end<<endl; if(T!=0) cout<<endl; } return 0;}
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