So Easy! HDU - 4565

​​So Easy! HDU - 4565​​

思路：题意虽然写着so easy 但是emmmm没想到怎么处理这个小数情况。看了题解：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
ll mod;
struct  matrix{
    ll x[2][2];
};
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]%=mod;
                if(temp.x[i][j]<0) temp.x[i][j]=(temp.x[i][j]+mod)%mod;//负数情况
            }
    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;
}
/*void p(matrix a)
{
    for(int i=0;i<5;i++)
    {
        for(int j=0;j<5;j++)
            printf("%lld,a.x[i][j]);
        printf("\n");
    }
}*/
int main()
{
    ll a,b,n;
    while(scanf("%lld%lld%lld%lld",&a,&b,&n,&mod)!=EOF)
    {
        matrix A;
        A.x[0][0]=(2*a)%mod;A.x[0][1]=(-(a*a)%mod+b)%mod;A.x[1][0]=1;A.x[1][1]=0;
        ll tmp=sqrt(b);
        if(tmp*tmp<b) tmp+=1;
        ll tmp2=2*a*sqrt(b);
        if(tmp2<2*a*tmp) tmp2+=1;
        tmp2=(a*a+b+tmp2)%mod;
        if(n<=2) n==1?printf("%lld\n",(a+tmp)%mod):printf("%lld\n",tmp2);
        else
        {
            matrix B;
            memset(B.x,0,sizeof(B.x));
            B.x[0][0]=(tmp2)%mod;B.x[1][0]=(a+tmp)%mod;
            A=quick_multi(A,n-2);
            B=multi(A,B);
            printf("%lld\n",(B.x[0][0]+mod)%mod);//负数情况
        }
    }
    return 0;
}