hdu3415 Max Sum of Max-K-sub-sequence

Max Sum of Max-K-sub-sequence
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit StatusDescription

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases.
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output a 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 minimum start position, if still more than one , output the minimum length of them.

Sample Input

4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1
Sample Output

7 1 3
7 1 3
7 6 2
-1 1 1

优先队列，在1-n加一个n-1就可以把环转化成一条线，用sum求合，那么从i到j的和就可以用sum[j]-sum[i],这个技巧也可以优化求和！然后把sum[i]用优先队列，j从0到n+m；这样一个一个求和，就可以了！

#include  
 #include  
using namespace std;
  int num[250000],sum[250000],prim[250000];  
int main()  {      int n,m,t,i,front,rear;   
   scanf("%d",&t); 
     while(t--)      {          scanf("%d%d",&n,&m);     
     sum[0]=0;     
     for(i=1;i<=n;i++)          {        
      scanf("%d",&num[i]);      
        sum[i]=sum[i-1]+num[i];//用合来简化运算        
  }          for(;i<=2*n;i++)   
       {             sum[i]=sum[i-1]+num[i-n];//大于Ｎ的部分i-n对应的相应的ＮＵＭ            }          front=0;     
     rear=0;      
    int maxx=-1e10,sx=0,ex=0;   
       for(i=1;i<=n+m;i++)   
       {              while(frontsum[i-1])//插入              {                  rear--;       
       }                prim[rear++]=i-1;        
      while(frontm)//去掉过界的           
   {                  front++;      
        }              if(maxxn)sx-=n;//注意大于n的其实是构造的模型，再重新  
        if(ex>n)ex-=n;      
    printf("%d %d %d\n",maxx,sx,ex);   
     }      return 0;  }