hdu3415 Max Sum of Max-K-sub-sequence(单调队列求n个数中和的最大值)

Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4836 Accepted Submission(s): 1765

Problem Description
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
题意：在组成环的n个数中求出不超过m个连续的数相加之和最大时的最大值、起始位置和终止位置(如果大于n则要减去n)
这个题我觉得还是有必要做一做的，虽然跟以前做的那个单调队列题有点像，可是这个求和的我还是第一次做

#include<stdio.h>   
int a[100005],sum[200005],q[200005];  
int main()  
{  
    int t,n,m,i,j,maxx,e,s,first,last;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d%d",&n,&m);  
        sum[0]=0;  
        for(i=1;i<=n;i++)  
        {  
            scanf("%d",&a[i]);  
            sum[i]=sum[i-1]+a[i];  
        }  
        for(;i<=n+m;i++)  
            sum[i]=sum[i-1]+a[i-n];//向后延伸m，这样就可以达到循环的效果   
        first=last=0;  
        q[last++]=0;//这要注意，WA好多次了，还是对单调队列理解的不够   
        maxx=a[1];  
        e=s=1;  
        for(i=1;i<=n+m;i++)  
        {  
            while(first<last&&sum[q[last-1]]>sum[i-1])//找最小值？因为我们看的是sum[i]-sum[q[first]],而sum[i]是定值   
                //所以要使这一项最大就必须使sum[q[first]]在m的范围内最小   
            {  
                last--;  
            }  
            q[last++]=i-1;//记录的也是i-1   
            while(first<last&&q[first]<i-m)//超出范围的去掉   
                first++;  
            if(maxx<sum[i]-sum[q[first]])  
            {  
                s=q[first]+1;//，我们当时记录的是i-1，这里要算起始点那就要加1   
                e=i;//终止点   
                maxx=sum[i]-sum[q[first]];//最大值   
            }  
        }  
        if(s>n)//这两句话千万别掉了   
            s-=n;  
        if(e>n)  
            e-=n;  
        printf("%d %d %d\n",maxx,s,e);  
    }  
    return 0;  
}

补充：软件开发 , C++ ,