【模板】矩阵快速幂

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题目

就是求一个矩阵的幂
快考试了时间太紧了没时间仔细搞
代码来自Qiu.YF

代码如下

#include<iostream>#include<cstdio>#include<cctype>#include<cstring>#include<string>    using namespace std;    #define in = read()    typedef long long ll;    typedef unsigned int ui;    const ll size = 100 +10;        ll n , k;        int mod = 1000000007;inline ll read(){        ll num = 0 , f = 1;    char ch = getchar();        while(!isdigit(ch)){                if(ch == '-')   f = -1;                ch = getchar();        }        while(isdigit(ch)){                num = num*10 + ch - '0';                ch = getchar();        }        return num*f;}struct Matrix{        ll s[size][size];        Matrix(){   memset(s , 0 , sizeof(s));}        Matrix operator *(const Matrix a){                Matrix ans;                for(register int i=1;i<=n;i++)                        for(register int j=1;j<=n;j++)                                for(register int k=1;k<=n;k++)                                        ans.s[i][j] = (ans.s[i][j] + s[i][k]*a.s[k][j]%mod)%mod;                return ans;        }};Matrix pow(Matrix q , ll opt){        ll i = opt;        Matrix ans;        for(register int j=1;j<=n;j++)  ans.s[j][j] = 1;        while(i){                if(i&1) ans = ans*q;                q = q*q;                i >>= 1;        }        return ans;}int main(){        n in;   k in;        Matrix start;        for(register int i=1;i<=n;i++)                for(register int j=1;j<=n;j++)                        start.s[i][j] in;        Matrix ans = pow(start , k);        for(register int i=1;i<=n;i++,printf("\n"))                for(register int j=1;j<=n;j++)                        printf("%lld " , ans.s[i][j]);        return 0;}//COYG

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