最大子段和
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 120365 Accepted Submission(s): 27835
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
//动规思想 #include<iostream> #define MIN -999999 using namespace std; int main() { int T,i,j,sum,n,cnt,start,end,a[100001],ans; cin>>T; cnt=0; while(cnt<T) { cin>>n; cnt++; ans = MIN; sum=0; j=1;//临时起始下标; for(i=1;i<=n;i++) { cin>>a[i]; sum+=a[i]; if(sum>ans){ans=sum;start=j;end=i;} if(sum<0){sum=0;j=i+1;} } cout<<"Case "<<cnt<<":"<<endl; cout<<ans<<" "<<start<<" "<<end<<endl; if(cnt!=T)cout<<endl; } system("pause"); return 0; }
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