HDU 4565 So Easy!

题目分析

Sn=⌈(a+b√)n⌉%m

因为(a−1)2<b<a2 ,所以0<(a−b√)n<1,因此Sn=⌈(a+b√)n+(a−b√)n⌉%m

接下来我们证明Sn=2∗a∗Sn−1+(b−a2)Sn−2。

Sn=(a+b√)n+(a−b√)n;Sn−1=(a+b√)n−1+(a−b√)n−1;移项得:Sn−(a−b√)nSn−1−(a−b√)n−1=a+b√;Sn−(a−b√)n=(a+b√)∗Sn−1−(a+b√)∗(a−b√)n−1;Sn=(a+b√)∗Sn−1−2∗a∗(a−b√)n−1;Sn−1=(a+b√)∗Sn−2−2∗a∗(a−b√)n−2;Sn−(a+b√)∗Sn−1Sn−1−(a+b√)∗Sn−2=a−b√化简得:Sn=2∗a∗Sn−1+(b−a2)∗Sn−2;

有上述的推导式子很明显可以推导出一个矩阵，然后利用矩阵快速幂求解即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL a,b,n,m;

struct Matirx{
int r,c;
LL matirx[2][2];
};

Matirx one, start, another;

Matirx Mul(Matirx a, Matirx b){  //矩阵乘法
Matirx c;
c.r = a.r;
c.c = b.c;
for(int i = 0; i < a.r; i++){
for(int j = 0; j < b.c; j++){
c.matirx[i][j] = 0;
for(int k = 0; k < a.c; k++)
c.matirx[i][j] += a.matirx[i][k]*b.matirx[k][j];
c.matirx[i][j] %= m;
}
}
return c;
}

Matirx Quick_pow(int n){     //矩阵快速幂
Matirx ret = one, temp = start;
while(n){
if(n&1) ret = Mul(ret, temp);
temp = Mul(temp, temp);
n >>= 1;
}
return ret;
}

void init(){ //初始化
one.r = one.c = start.c = start.r = 2;
one.matirx[0][0] = 1;
one.matirx[0][1] = 0;
one.matirx[1][0] = 0;
one.matirx[1][1] = 1;
start.matirx[0][0] = 0;
start.matirx[0][1] = 1;
start.matirx[1][0] = b-a*a;
start.matirx[1][1] = 2*a;
another.r = 2;
another.c = 1;
another.matirx[0][0] = 2*a;
another.matirx[1][0] = 2*(a*a+b);
}

int main(){
while(scanf("%I64d%I64d%I64d%I64d", &a, &b, &n, &m) != EOF){
init();
if(n == 1) printf("%I64d\n", (2*a)%m);
else if(n == 2) printf("%I64d\n", (2*(a*a+b))%m);
else {
Matirx ret = Quick_pow(n-2);
another = Mul(ret, another);
printf("%I64d\n", (another.matirx[1][0]+m)%m); //这里一定要注意,因为构建start矩阵时你会发现b-a*a为负数，因此可能会出现负值
}
}
return 0;
}