Max Sum of Max-K-sub-sequence HDU

题目链接:点我

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    Given a circle sequence A11,A22,A33......Ann. Circle sequence means the left neighbour of A11 is Ann , and the right neighbour of Ann is A11.     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

46 36 -1 2 -6 5 -56 46 -1 2 -6 5 -56 3-1 2 -6 5 -5 66 6-1 -1 -1 -1 -1 -1

Sample Output

7 1 37 1 37 6 2-1 1 1

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题意:

求大小为k以内的子序列的最大和.

思路:

序列倍增,然后求前缀和,最后用前缀和维护一个下标差不超过m的单调递增的单调队列.

代码:

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>using namespace std;const int maxn = 200000 + 10;int sum[maxn];int a[maxn];int b[maxn];int main(){    int t;    scanf("%d", &t);    while(t--){        int  n, m;        scanf("%d %d", &n, &m);        for(int i = 1;i <= n;++i){            scanf("%d", &a[i]);            a[n+i]=a[i];        }        sum[0] = 0;        int ansl,ansr,ans=-100000000;        for(int  i = 1; i <= n + n;++i){//找出序列中最大的数并且求前缀和            sum[i] = sum[i-1] + a[i];            if(a[i] > ans){                ans = a[i];                ansl =i;                ansr = i;            }        }        int l= 0 ,r ,k = 0;        for(int i = 1; i <= n + n;++i){            while( k > l && sum[b[k-1]]>sum[i-1]) k--;            b[k++] = i-1;           while(k > l && i - b[l] > m ) ++l;//单调队列的大小不超过m            if((ans <sum[i] -  sum[b[l]] && i)){                ans = sum[i] - sum[b[l]];                ansl = b[l] + 1;                ansr = i;            }        }        if (ansl > n) ansl -= n;        if (ansr > n) ansr -= n;        printf("%d %d %d\n",ans ,ansl, ansr);    }    return 0;}