Poj 3233

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题目

类似于等比数列求和,用二分.

一开始取余符号多用了几次,直接TLE了==

n%2==0: F[n]=F[n/2] (1+a[n/2]);

n%2==1: F[n]=F[n-1]+a[n];

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,mod,k;struct Matrix{    int m[31][31];}E,Z;inline Matrix Mut(Matrix A,Matrix B){    Matrix ans;    for(int i=0;i<n;i++)    for(int j=0;j<n;j++)    {        ans.m[i][j]=0;        for(int k=0;k<n;k++)        {            ans.m[i][j]+=((A.m[i][k])*(B.m[k][j]));            ans.m[i][j]%=mod;        }    }    return ans;}inline Matrix Add(Matrix A,Matrix B){    Matrix ans;    for(int i=0;i<n;i++)    for(int j=0;j<n;j++)    ans.m[i][j]=(A.m[i][j]+B.m[i][j])%mod;    return ans;}inline Matrix Pow(Matrix A,int b){    Matrix t=A,ans=E;    while(b)    {        if(b%2) ans=Mut(ans,t);        b/=2;        t=Mut(t,t);    }    return ans;}inline Matrix solve(Matrix A,int b){    if(b==0) return E;    else if(b==1) return A;    else if(b==2) return Add(A,Mut(A,A));    else if(b%2==1) return Add(Pow(A,b),solve(A,b-1));    else if(b%2==0)    {        Matrix ans=solve(A,b/2);        Matrix C=Pow(A,b/2);        Matrix B=Mut(ans,C);        return Add(B,ans);    }}int main(){    //freopen("in.txt","r",stdin);    scanf("%d%d%d",&n,&k,&mod);    {        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        E.m[i][j]=(i==j);        Matrix A;        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        scanf("%d",&A.m[i][j]),A.m[i][j]%=mod;        Matrix ans=solve(A,k);        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(j!=0)                printf(" ");                printf("%d",ans.m[i][j]%mod);            }            printf("\n");        }    }    return 0;}