大哥们帮看下 郁闷呀。杭电ACM的题。
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49751 Accepted Submission(s): 11077
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
要求each line starts with a number N(1<=N<=100000) 这怎么实现啊？ 我写的又通不过。
# include <iostream>using namespace std;int main(){ int i,j,k; //control circlation int a[20][100000],b[20][4]; int term; int sum=0,sum1=0; cin>>term; for(i=0;i<term;i++) { cin>>b[i][0]; for(j=0;j<b[i][0];j++) { cin>>a[i][j]; } } for(k=0;k<term;k++) { for(i=0;i<=b[k][0];i++) { for(j=i;j<=b[k][0];j++) { sum+=a[k][j]; if(sum>b[k][1]) { b[k][1]=sum; b[k][2]=i; b[k][3]=j; } } sum=0; } } for(i=0;i<term;i++) { cout<<"Case"<<" "<<i+1<<":"<<endl; cout<<b[i][1]<<ends<<b[i][2]+1<<ends<<b[i][3]+1<<endl; cout<<endl; } return 0;}