srm#397_div1_500pt 矩阵乘法+快速模幂

题目地址：srm#397_div1_500

题目描述：

Problem Statement

NOTE: This problem statement contains superscripts that may not display properly if viewed outside of the applet.

You are given ints n and k. Return the value of the sum 1^k + 2^k + 3^k + ... + n^k modulo 1000000007.

Definition

Class:	SumOfPowers
Method:	value
Parameters:	int, int
Returns:	int
Method signature:	int value(int n, int k)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	64

Constraints

n will be between 1 and 10⁹, inclusive.

k will be between 1 and 50, inclusive.

Examples

Returns: 15

Here, we have arithmethic progression: 1 + 2 + 3 + 4 + 5 = 15.

Returns: 30

Just a little bit more complicated example here: 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30.

Returns: 1002001

This one would be harder to check by hand.

123456789

Returns: 383478132

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构造一个k+2阶矩阵；

这里n比较大显然我们接受不了O（n）的复杂度

最好的方法就是像求Fibonacci数列第n项一样用矩阵乘法+快速幂做

要实现求和长度在k左右的递推式

首选（n+1）^k-n^k=sigma(c[k][i]*n^i);

直接看代码里面的矩阵构造吧

wa了几次

1 组合数是要求到c[52][i]的

2 因为mod 10^9+7 所以存在的数都是可能接近int上限的要进行+运算所以要用long long 存储

代码：

#include<iostream>typedef  long long inta ;using namespace std;struct Matrix
{inta  m[60][60];};inta n;    //  用来表示维度inta   c[60][60];    //组合数const   inta  mod=1000000007;void init()
{for(inta i=0;i<=55;i++)c[i][0]=1;for(inta i=0;i<55;i++)for(inta j=0;j<=i;j++)c[i+1][j+1]=c[i][j+1]+c[i][j];}Matrix multi(Matrix a,Matrix b)
{Matrix ans;for(inta i=0;i<n;i++)for(inta j=0;j<n;j++){inta   c=0;for(inta k=0;k<n;k++)c=(c+a.m[i][k]*b.m[k][j])%mod;ans.m[i][j]=c;}return ans;}Matrix quick_mod(Matrix a,inta b)
{Matrix  ans;Matrix  p=a;for(inta i=0;i<n;i++)for(inta j=0;j<n;j++)ans.m[i][j]=(i==j?1:0);while(b){if(b&1){ans=multi(ans, p);b--;}b>>=1;p=multi(p, p);}return ans;}class  SumOfPowers
{public :inta value(inta nn,inta k){init();n=k+2;Matrix   A;for(inta i=0;i<n;i++)      //  so  importantfor(inta j=0;j<n;j++)A.m[i][j]=0;for(inta i=0;i<k+1;i++)for(inta j=0;j<=i;j++)A.m[i][j]=c[i][j]%mod;for(inta i=0;i<k+1;i++)A.m[k+1][i]=c[k][i]%mod;A.m[k+1][k+1]=1;A=quick_mod(A, nn-1);inta   ans=0;for(inta i=0;i<k+2;i++)ans=(ans+A.m[k+1][i])%mod;return ans;}
};int  main()
{inta  nn,k;cin>>nn>>k;SumOfPowers  obj;cout<<obj.value(nn, k)<<endl;}

tc上提交没有main()

转载于:https://www.cnblogs.com/jingqi814/p/3644368.html