Reading comprehension HDU - 4990
Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
const int MAX=100000*2;
const int INF=1e9;
int main()
{
int n,m,ans,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(i=1;i<=n;i++)
{
if(i&1)ans=(ans*2+1)%m;
else ans=ans*2%m;
}
printf("%d\n",ans);
}
return 0;
}
Input
Multi test cases,each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000
Output
For each case,output an integer,represents the output of above program.
Sample Input
1 10
3 100
Sample Output
1
5
题意:
用更快速的方法求上述代码
分析:
通过上面的代码:
我们知道了对于
更希望得到一个可以用一个式子表示的递推式,这样就可以用矩阵快速幂快速求解
通过打表可以得到
因此得到关系矩阵
code:
//a(n) = a(n-1) + 2*a(n-2) + 1
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m;
struct Matrix{
ll mat[4][4];
Matrix operator * (const Matrix &b)const{
Matrix ans;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
ans.mat[i][j] = 0;
for(int k = 0; k < 3; k++){
ans.mat[i][j] += mat[i][k] * b.mat[k][j] % m;
ans.mat[i][j] %= m;
}
}
}
return ans;
}
};
Matrix q_pow(Matrix a,ll b){
Matrix ans;
memset(ans.mat,0,sizeof(ans.mat));
for(int i = 0; i < 3; i++){
ans.mat[i][i] = 1;
}
while(b){
if(b & 1)
ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
int main(){
while(~scanf("%lld%lld",&n,&m)){
Matrix ans;
ans.mat[0][0] = ans.mat[0][2] = ans.mat[1][0] = ans.mat[2][2] = 1;
ans.mat[1][1] = ans.mat[1][2] = ans.mat[2][0] = ans.mat[2][1] = 0;
ans.mat[0][1] = 2;
ans = q_pow(ans,n-1);
ll ret = (ans.mat[0][0] + ans.mat[0][2]) % m;
printf("%lld\n",ret);
}
return 0;
}