ZOJ 2105 Number Sequence(矩阵快速幂)

注: 第一想法就是矩阵快速幂，还有一想法是maybe有规律可循，这里用的是矩阵快速幂解法

Number Sequence

Time Limit: 2 Seconds Memory Limit: 65536 KB

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

构建矩阵:(矩阵竖线没那么长，将就看下吧)



发现规律了吧，矩阵快速幂模板

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#define N 2
#define mod 7
typedef long long LL;
using namespace std;
struct Mar{
int a

;
void init(){
for(int i=0;i<2;i++)
a[i][i]=1;
}
void zero(){
memset(a,0,sizeof(a));
}
};
Mar operator * (const Mar x,const Mar y){
Mar t;
t.zero();
for(int k=0;k<2;k++){
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
t.a[i][j]+=x.a[i][k]*y.a[k][j];
t.a[i][j]%=mod;
}
}
}
return t;
}
Mar operator ^ (Mar x,int n){
Mar t;
t.zero();
t.init();
while(n){
if(n%2==1) t=t*x;
x=x*x;
n/=2;
}
return t;
}
int main(){
int n,a,b,ans;
Mar t;
while(scanf("%d%d%d",&a,&b,&n)!=EOF){
if(!n&&!a&&!b) break;
if(n==1) ans=1;
else if(n==2) ans=1;
else{
t.a[0][0]=a,t.a[0][1]=1,t.a[1][0]=b,t.a[1][1]=0;
t = t^(n-2);
ans=t.a[0][0]+t.a[1][0];
}
printf("%d\n",ans%mod);
}
return 0;
}