Yet another Number Sequence UVA - 10689

​​Yet another Number Sequence UVA - 10689​​​
题意：给出a,b,n,m。f(0)=a,f(1)=b,求f(n)=f(n-1)+f(n-2)的最后m位数字。
思路：矩阵的模板题。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
struct  matrix{
    ll x[2][2];
};
ll m;
matrix multi(matrix a,matrix b){
    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]%=m;//负数取模的问题,除法取模
            }
    return temp;
}
matrix quick_multi(matrix a,ll n)//矩阵快速幂
{
    matrix temp=a;
    n--;
    while(n){
        if(n&1)
            temp=multi(temp,a);
        a=multi(a,a);
        n>>=1;
    }
    return temp;
}
int main()
{
    int t;
    scanf("%d",&t);
    ll a,b,n;
    while(t--)
    {
        scanf("%lld%lld%lld%lld",&a,&b,&n,&m);
        int mod=m;m=1;
        while(mod--) m*=10;
        matrix A,B;
        memset(A.x,0,sizeof(A.x));
        memset(B.x,0,sizeof(B.x));
        A.x[0][0]=A.x[0][1]=A.x[1][0]=1;
        B.x[0][0]=b,B.x[1][0]=a;
        if(n<=1) n==0?printf("%d\n",a%m):printf("%d\n",b%m);
        else
        {
            A=quick_multi(A,n-1);
            B=multi(A,B);
            printf("%d\n",B.x[0][0]%m);
        }
    }
    return 0;
}