HDU1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 113998    Accepted Submission(s): 26377

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

Author

Ignatius.L

 

O(n)算法：

#include<iostream>#include<stdio.h>#define maxn 100004using namespace std;int a[maxn];int main(){    int n,m;   cin>>n;   int count=0;   while(n--)   {       count++;       cin>>m;       int sum=0;       int l,l_o=1,r;       int MAX=-1005;       int flag=0;       for(int i=1;i<=m;i++)       {           cin>>a[i];           if(a[i]>0) flag=1;           sum+=a[i];           if(MAX<sum) {MAX=sum;r=i;l=l_o;}//l=l_o记录上一个连续和的首位           if(sum<0)           {               sum=0;               l_o=i+1;           }        }       if(!flag)       {           MAX=a[1];           for(int i=2;i<=m;i++)           {              if(MAX<a[i]) {MAX=a[i];r=l=i;}//只有一个数输出首尾位置数字相同           }       }       cout<<"Case "<<count<<":"<<endl;//因为格式问题WA数次       cout<<MAX<<" "<<l<<" "<<r<<endl;       if(n) cout<<endl;//注意最后一组数据不输出换行   }   return 0;}