hdu 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): 112168 Accepted Submission(s): 25902
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 -5Sample OutputCase 1:14 1 4Case 2:7 1 6
#include<cstdio>#include<iostream>#include<string.h>#include<iostream>using namespace std;int main(){ int T,icase=1; cin>>T; while(T--) { int a[100005],N,i,k,front,rear,sum=0,max=-111111111; cin>>N; for(i=0;i<N;i++) cin>>a[i]; for(i=0,k=0;i<N;i++) { sum+=a[i]; if(max<sum){max=sum;front=k,rear=i;} if(sum<0) {sum=0;k=i+1;} } printf("Case %d:\n",icase++); printf("%d %d %d\n",max,front+1,rear+1); if(T) printf("\n"); } return 0;}
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