COJ 2124 Day8-例1

Day8-例1

难度级别：B； 运行时间限制：1000ms； 运行空间限制：256000KB； 代码长度限制：2000000B

试题描述

给定n、m的值，求

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答案对10^9+7取模。

输入

一行，两个整数n、m。

输出

一行，一个整数，表示答案mod 10^9+7的值。

输入示例

5 3

输出示例

36363

其他说明

n<=10^9 m<=50

题解：矩阵乘法+变形+二项式定理：

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<stack>
 #include<queue>
 #include<cstring>
 #define PAU putchar(' ')
 #define ENT putchar('\n')
 using namespace std;
 typedef long long LL;
 const LL mod=;
 const int maxn=+;
 struct matx{
     int h,w;LL x[maxn][maxn];matx(){memset(x,,sizeof(x));}
     matx operator*(matx a){
         matx b;b.h=h;b.w=a.w;
         for(int k=;k<w;k++)
             for(int i=;i<h;i++)
                 for(int j=;j<a.w;j++)b.x[i][j]=(b.x[i][j]+x[i][k]*a.x[k][j])%mod;return b;
     }
     void init(int h,int w){this->h=h;this->w=w;return;}
 }A,B;
 LL C[maxn][maxn];int n,m;
 LL powi(LL x,LL y){
     LL ans=;x%=mod;for(int i=y;i;i>>=,x=x*x%mod)if(i&)ans=ans*x%mod;return ans;
 }
 matx powm(matx a,int n){
     matx ans=a,tmp=a;n--;for(int i=n;i;i>>=,tmp=tmp*tmp)if(i&)ans=ans*tmp;return ans;
 }
 LL inv(LL a){return powi(a,mod-);}
 void initc(){
     for(int i=;i<=m;i++)C[i][]=,C[i][i]=;
     for(int i=;i<=m;i++)for(int j=;j<i;j++)C[i][j]=(C[i-][j]+C[i-][j-])%mod;return;
 }
 void initmatx(){
     A.init(,m+);B.init(m+,m+);A.x[][]=;
     for(int i=;i<=m;i++)for(int j=;j<=i;j++)B.x[j][i]=C[i][j];
     for(int i=;i<=m;i++)B.x[i][m+]=C[m][i];
     B.x[m+][m+]=inv(m);return;
 }
 inline int read(){
     int x=,sig=;char ch=getchar();
     for(;!isdigit(ch);ch=getchar())if(ch=='-')sig=;
     for(;isdigit(ch);ch=getchar())x=*x+ch-'';
     return sig?x:-x;
 }
 inline void write(long long x){
     if(x==){putchar('');return;}if(x<)putchar('-'),x=-x;
     int len=;long long buf[];while(x)buf[len++]=x%,x/=;
     for(int i=len-;i>=;i--)putchar(buf[i]+'');return;
 }
 int main(){
     n=read();m=read();initc();initmatx();A=powm(B,n);
     write((A.x[][m+]*powi(m,n))%mod);
     return ;
 }