hdu-6063-RXD and math

最新推荐文章于 2018-07-19 09:14:06 发布

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最新推荐文章于 2018-07-19 09:14:06 发布

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探讨数学家RXD面临的复杂数学计算问题，包括求和公式∑i=1nkμ²(i)×⌊nki−−−√⌋的计算方法，并提供了一种高效的算法实现，该算法使用模块化指数运算来解决大规模数值计算问题。

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题目链接

Problem Description

RXD is a good mathematician.
One day he wants to calculate:

\sum i = 1 n k μ 2 (i) \times ⌊ n k i - - - \sqrt ⌋

output the answer module

109+7 .

1≤n,k≤1018

μ (n) = 1 (n = 1)

μ (n) = (- 1) k (n = p 1 p 2 \dots p k)

μ (n) = 0 (o t h e r w i s e)

p1,p2,p3…pk are different prime numbers

Input

There are several test cases, please keep reading until EOF.
There are exact 10000 cases.
For each test case, there are 2 numbers

n,k .

Output

For each test case, output "Case #x: y", which means the test case number and the answer.

Sample Input

  
   10 10

Sample Output

  
   Case #1: 999999937

Source

规律：求 n的k次方然后取模，别问我怎么知道的，我是猜的。摊手、、、、、、

代码：

#include<bits/stdc++.h>
#define mod 1000000007
#define mod 1e9+7
using namespace std;
long long pow(long long a,long long b)
{
   long long int ans = 1,base = a;
    while(b!=0)
    {
        if(b&1){
            base%=(long long)mod;
            ans%=(long long)mod;
            ans *= base;
            ans%=(long long)mod;
        }
        base%=(long long)mod;
        base *= base;
        base%=(long long)mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    long long int n,k;
    int t=0;
    while(~scanf("%lld %lld",&n,&k)){
            t++;
        printf("Case #%d: ",t);
        printf("%lld\n",pow(n,k));
    }
    return 0;
}