bzoj2875: [Noi2012]随机数生成器

没改LL见祖宗系列。

矩乘不难推吧。然后要用快速乘

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long LL;
LL mod;

LL n,AA,C,X,G;
struct Matrix
{
    LL a[5][5];
    Matrix(){}
    Matrix clear()
    {
        memset(a,0,sizeof(a));
    }
};

LL quick_multi(LL k,LL p)
{
    if(k==0||p==0)return 0LL;
    LL A=k,ret=0;
    while(p!=0)
    {
        if(p%2==1)ret=(ret+A)%mod;
        A=(A+A)%mod;p/=2;
    }
    return ret;
}
Matrix multi(Matrix A,Matrix B)
{
    Matrix C;C.clear();
    for(int i=1;i<=2;i++)    
        for(int j=1;j<=2;j++)
            for(int k=1;k<=2;k++)
                C.a[i][j]=(C.a[i][j]+quick_multi(A.a[i][k],B.a[k][j]))%mod;
    return C;                
}
Matrix quick_pow(LL p)
{
    Matrix A,ret;A.clear();ret.clear();
    A.a[1][1]=1;A.a[1][2]=0;
    A.a[2][1]=1;A.a[2][2]=AA;
    ret.a[1][1]=1;ret.a[1][2]=0;
    ret.a[2][1]=0;ret.a[2][2]=1;
    while(p!=0)
    {
        if(p%2==1)ret=multi(ret,A);
        A=multi(A,A);p/=2;
    }
    return ret;
}
int main()
{
    scanf("%lld%lld%lld%lld%lld%lld",&mod,&AA,&C,&X,&n,&G);
    Matrix ans;ans.clear();
    ans.a[1][1]=C;ans.a[2][1]=X;
    ans=multi(quick_pow(n),ans);
    printf("%lld
",ans.a[2][1]%G);
    return 0;
}
原文地址:https://www.cnblogs.com/AKCqhzdy/p/8643490.html