UVA - 10689 Yet another Number Sequence

题目链接

简单的矩阵快速幂问题，取n的最后m位数字即为模取10的m次方。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
//const ll mod=1e9+7;
ll mod;
struct matrix{
	ll x[2][2];
};
matrix multi(matrix a,matrix b)//矩阵相乘
{
	matrix temp;
	memset(temp.x,0,sizeof(temp.x));
	for(int i=0;i<2;i++)
		for(int j=0;j<2;j++)
			for(int k=0;k<2;k++)
			{
				temp.x[i][j]+=a.x[i][k]*b.x[k][j];
				temp.x[i][j]%=mod;
			}
	return temp;
}
matrix quick_multi(matrix a,ll k)//矩阵快速幂
{
	matrix temp=a;
	k--;
	while(k){
		if(k&1)
			temp=multi(temp,a);
		a=multi(a,a);
		k>>=1;
	}
	return temp;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--){
		int a,b,m;
		ll n;
		scanf("%d%d%lld%d",&a,&b,&n,&m);
		mod=1;
		while(m--) mod*=10;
		if(n==0){
			printf("%d\n",a%mod);
			continue;
		}
		if(n==1){
			printf("%d\n",b%mod);
			continue;
		}
		matrix A;
		matrix ans;
		memset(A.x,0,sizeof(A.x));
		memset(ans.x,0,sizeof(ans.x));
		A.x[0][0]=1;A.x[1][0]=1;
		A.x[0][1]=1;A.x[1][1]=0;
		
		ans.x[0][0]=b;ans.x[0][1]=a;
		
		A=quick_multi(A,n-1);
		ans=multi(ans,A);
		printf("%lld\n",(ans.x[0][0]+mod)%mod);
			
	} 
	return 0;
}