轉載:http://www.klogk.com/posts/hdu4565/
這裡寫的很好,看看就知道了啊。
題意非常簡單,a,b,n都是正整數,求
Sn=⌈(a+b√)n⌉%m,(a−1)2<b<a2
這個題目也是2008年Google Codejam Round 1A的C題。
做法其實非常簡單,記(a+b√)n為An,配項
Cn=An+Bn=(a+b√)n+(a−b√)n
兩項恰好共軛,所以Cn是整數。又根據限制條件
(a−1)2<b<a2⇒0<a−b√<1⇒0<(a−b√)n<1⇒Bn<1
也就是說Cn=⌈An⌉
Sn=(Cn)%m
求Cn的方法是遞推。 對Cn乘以(a+b√)+(a−b√)
Cn[(a+b√)+(a−b√)]=[(a+b√)n+(a−b√)n][(a+b√)+(a−b√)]=(a+b√)n+1+(a−b√)n+1+(a+b√)n(a−b√)+(a−b√)n(a+b√)=Cn+1+(a2−b)(a+b√)n−1+(a2−b)(a−b√)n−1=Cn+1+(a2−b)Cn−1
於是
Cn+1=2aCn−(a2−b)Cn−1
把這個遞推式寫成矩陣形式
[Cn+1Cn]=[2a1−(a2−b)0][CnCn−1]
於是就可以用矩陣快速冪來做了
[Cn+1Cn]=[2a1−(a2−b)0]n[C1C0]
So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2286 Accepted Submission(s): 710
Problem Description
A sequence Sn is defined as:
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
Input
There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the end of file.
Output
For each the case, output an integer Sn.
Sample Input
2 3 1 2013
2 3 2 2013
2 2 1 2013
Sample Output
4
14
4
<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
///#define mod 10007
const int maxn = 210;
using namespace std;
struct matrix
{
LL f[3][3];
};
LL mod;
matrix mul(matrix a, matrix b, int n)
{
matrix c;
memset(c.f, 0, sizeof(c.f));
for(int k = 0;k < n; k++)
{
for(int i = 0; i < n;i++)
{
if(!a.f[i][k]) continue;
for(int j = 0; j < n; j++)
{
if(!b.f[k][j]) continue;
c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j]+mod)%mod;
}
}
}
return c;
}
matrix pow_mod(matrix a, LL b, int n)
{
matrix s;
memset(s.f, 0 , sizeof(s.f));
for(int i = 0; i < n; i++) s.f[i][i] = 1LL;
while(b)
{
if(b&1) s = mul(s, a, n);
a = mul(a, a, n);
b >>= 1;
}
return s;
}
int main()
{
LL a, b, n;
while(~scanf("%I64d %I64d %I64d %I64d",&a, &b, &n, &mod))
{
if(n == 1)
{
printf("%I64d\n",2*a%mod);
continue;
}
matrix c;
memset(c.f, 0 , sizeof(c.f));
c.f[0][0] = 2LL*a;
c.f[0][1] = b-a*a;
c.f[1][0] = 1LL;
matrix d = pow_mod(c, n-1, 2);
printf("%I64d\n",((d.f[0][0]*2*a+d.f[0][1]*2)%mod+mod)%mod);
}
return 0;
}</span>