hdu 5793 A Boring Question （快速幂+逆元）(规律题)

最新推荐文章于 2021-05-27 11:38:34 发布

ailincheng0815

最新推荐文章于 2021-05-27 11:38:34 发布

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原文链接：http://www.cnblogs.com/yu0111/p/5740013.html

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本文介绍了一种计算等比数列前n+1项和的高效算法，通过快速幂运算和逆元求解，实现了对大规模数据的快速处理。代码采用C++实现，适用于竞赛编程和大数据处理场景。

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官方推导公式：

$\sum 0 \leq k 1 , k 2 , \dots k m \leq n \prod 1 \leq j < m ( k j k j + 1 )$ $=\sum_{0\leq k_{1}\leq k_{2}\leq\cdots \leq k_{m}\leq n}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}}$

$= \sum 0 \leq k 1 \leq k 2 \leq \dots \leq k m \leq n \prod 1 \leq j < m ( k j k j + 1 )$ $=\sum_{k_{m}=0}^{n}\sum_{k_{m-1}=0}^{k_{m}}\cdots \sum_{k_{1}=0}^{k_{2}}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}}$

$= \sum k m = 0 n \sum k m - 1 = 0 k m \dots \sum k 1 = 0 k 2 \prod 1 \leq j < m ( k j k j + 1 )$ $=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{1}=0}^{k_{2}}\binom{k_{2}}{k_{1}} \right } \right }$

$= \sum k m = 0 n {( k m - 1 k m ) \sum k m - 1 = 0 k m {( k m - 2 k m - 1 ) \dots \sum k 1 = 0 k 2 ( k 1 k 2 )}}$

$=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{1}=0}^{k_{2}}\binom{k_{2}}{k_{1}} \right } \right }$ $=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{2}=0}^{k_{3}}\binom{k_{3}}{k_{2}}2^{k_{2}} \right } \right }$

$= \sum k m = 0 n {( k m - 1 k m ) \sum k m - 1 = 0 k m {( k m - 2 k m - 1 ) \dots \sum k 2 = 0 k 3 ( k 2 k 3 ) 2 k 2 }}$ $=\sum_{k_{m}=0}^{n}m^{k_{m}}$

$= \sum k m = 0 n m k m $ $=\frac{m^{n+1} - 1}{m - 1}$

$= m - 1 m n + 1 - 1 $

自己找的规律是求：ans=m^0+m^1+m^2+...+m^n，也就是等比数列的前n+1项和

/*by*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const LL N=2010;
const LL mod=1000000007;
const LL INF=0x3f3f3f;
LL quickpow(LL m, LL n)
{
    LL b = 1;
    while(n > 0) {
        if(n & 1)
            b = (b * m) % mod;
        n = n >> 1 ;
        m = (m * m) % mod;
    }
    return b;
}
int main()
{
    int T;
    cin>>T;
    while(T--) {
        LL n,m;
        scanf("%lld%lld",&n,&m);
        LL ans=0;
        ans=quickpow(m,n+1)-1;
        ans=ans*quickpow(m-1,mod-2)%mod;/*求逆元*/
        printf("%lld\n",ans%mod);
    }
    return 0;
}