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自然万物都趋向从有序变得无序

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Yet another Number Sequence UVA - 10689 (矩阵快速幂)

题目链接:点我

题目


题意:

求f(n) mod m 的值,

思路:

类似与求斐波那契数列(Fibonacci)数列的第n项那样,快速幂即可.
构造递推矩阵为:$ \begin{pmatrix} 1 & 1 \ 1 & 0\end{pmatrix}$

代码:

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;

int mod[]={1,10,100,1000,10000};
int m;

struct mat{
int a[10][10];
mat(){memset(a, 0, sizeof(a)); }
mat operator *(const mat q){
mat c;
for(int i = 1; i <= 2; ++i)
for(int k = 1; k <= 2; ++k)
if(a[i][k])
for(int j = 1; j <=2; ++j){
c.a[i][j] += a[i][k] * q.a[k][j];
if(c.a[i][j] >= mod[m]) c.a[i][j] %= mod[m];
}return c;
}
};

mat qpow(mat x, int n){
mat ans;
ans.a[1][1] = ans.a[2][2] = 1;
while(n){
if(n&1) ans = ans * x;
x= x * x;
n >>= 1;
}return ans;
}

int main(){
int a,b,n;
int t;
scanf("%d", &t);
while(t--){
scanf("%d %d %d %d", &a, &b, &n, &m);
a %= mod[m];
b %= mod[m];
if(n == 0){
printf("%d\n",a);
continue;
} else if(n == 1){
printf("%d\n",b);
continue;
}mat ans;
ans.a[1][1] = ans.a[1][2] = ans.a[2][1] = 1;
ans = qpow(ans,n-1);
printf("%d\n",(ans.a[1][1] * b + ans.a[1][2] * a) % mod[m]);
}return 0;
}