bzoj2875: [Noi2012]随机数生成器

题目链接

bzoj2875: [Noi2012]随机数生成器

题解

递推式有了，那就愉快的矩阵快速幂

代码

/* 
递推式子都给你了，那就愉快的矩阵快速幂呗 
*/ 
#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define LL long long
LL m,a,c,x0,n,g; 
LL mul(LL x, LL y) { 
    LL ret = 0; 
    for(;y;y >>= 1, x = (x + x) % m) 
        if(y & 1) ret = (ret + x) % m; 
    return ret % m; 
} 
struct Matrix { 
	LL a[3][3]; 
	int r,c; 
	Matrix() {memset(a,0,sizeof a);r = 0,c = 0;}  
	Matrix operator * (const Matrix &A) const { 
		Matrix ret; ret.r = r,ret.c = A.c; 
		for(int i = 0;i < r;++ i) 
			for(int j = 0;j < A.c;++ j) 
				for(int k = 0;k < c;++ k) 
					ret.a[i][j] = (ret.a[i][j] + mul(a[i][k] , A.a[k][j]) ) % m;  
		return ret; 
	} 
} f; 
Matrix fstpow() { 
	Matrix ret; 
	ret.r = 2,ret.c = 2; 
	ret.a[0][0] = 1,ret.a[1][1] = 1; 
	for(;n;n >>= 1ll,f = f * f) 
		if(n & 1ll) ret = f * ret; 
	return ret;  
} 
int main() { 
	scanf("%lld%lld%lld%lld%lld%lld",&m,&a,&c,&x0,&n,&g); 
	f.r = 2,f.c = 2; 
	f.a[0][0] = a;f.a[1][1] = f.a[0][1] = 1; 
	Matrix ans; ans.r = 2,ans.c = 1; 
	ans.a[0][0] = x0; ans.a[1][0] = c; 
	Matrix tmp = fstpow(); 
	ans = tmp * ans; 
	printf("%lld\n",ans.a[0][0] % g);  
	return 0;  
}