2017杭电多校联赛team3 Kanade's sum hdu6058 快速幂

Kanade's sum

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2011    Accepted Submission(s): 810

Problem Description

Give you an array A[1..n]of length n. 

Let f(l,r,k) be the k-th largest element of A[l..r].

Specially , f(l,r,k)=0 if r−l+1<k.

Give you k , you need to calculate ∑nl=1∑nr=lf(l,r,k)

There are T test cases.

1≤T≤10

k≤min(n,80)

A[1..n] is a permutation of [1..n]

∑n≤5∗105

 

Input

There is only one integer T on first line.

For each test case,there are only two integers n,k on first line,and the second line consists of n integers which means the array A[1..n]

 

Output

For each test case,output an integer, which means the answer.

 

Sample Input

15 21 2 3 4 5

 

Sample Output

30

 

Source

2017 Multi-University Training Contest - Team 3

 

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思路：

由样例发现，n^k%mod=ans。

官方题解

ac代码：

#include<stdio.h>#define LL long long#define Mod 1e9+7LL N,K,ans;LL power2(LL a, LL b, LL c) {LL res = 1;a %= c;while (b) {if (b & 1)res = (res * a) % c;a = (a * a) % c;b >>= 1;}return res;}int main() {int cnt=0;while(~scanf("%lld%lld",&N,&K)) {printf("Case #%d: %lld\n",++cnt,power2(N,K,Mod));}return 0;}

总结：有想法多试一试，数据大的题目打表看看（上一场多校貌似也有一道找规律的）。