版权声明:有错误欢迎大家指出。转载请注明出处~ https://blog.csdn.net/black_miracle/article/details/76474362
Funny Function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1215 Accepted Submission(s): 596
Problem Description
Function
Fx,ysatisfies:
For given integers N and M,calculate
Fm,1 modulo 1e9+7.
Input
There is one integer T in the first line.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.
Output
For each given N and M,print the answer in a single line.
Sample Input
2 2 2 3 3
Sample Output
2 33
题意:求f m 1
题解:打表找到规律。。。
然后矩阵快速幂一步一步一步一步一步一步去完成
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
ll N;
struct Matrix{
ll a[4][4];
};
Matrix mul(Matrix a,Matrix b){
Matrix c;
memset(c.a,0,sizeof(c.a));
for(ll i=1;i<=N;i++){
for(ll j=1;j<=N;j++){
if(!a.a[i][j])continue;
for(ll k=1;k<=N;k++){
c.a[i][k]+=a.a[i][j]*b.a[j][k];
c.a[i][k]%=mod;
}
}
}
return c;
}
Matrix quick(Matrix a,ll k){
Matrix ans;
memset(ans.a,0,sizeof(ans));
for(ll i=1;i<=N;i++)ans.a[i][i]=1;
while(k>0){
if(k&1)ans=mul(ans,a);
a=mul(a,a);
k/=2;
}
return ans;
}
ll qq(ll a,ll k){
ll ans=1;
while(k){
if(k&1)ans=ans*a%mod;
a=a*a%mod;
k/=2;
}
return ans;
}
int main(){
int t,cas=1;
scanf("%d",&t);
while(t--){
ll n,m;
scanf("%lld%lld",&n,&m);
if(m==1){
printf("1\n");
continue;
}
Matrix f2;
memset(f2.a,0,sizeof(f2.a));
f2.a[1][1]=f2.a[1][2]=1;
f2.a[2][1]=2;
N=2;
f2=quick(f2,n-1);
ll ps=f2.a[1][1]+f2.a[2][1];
if(n%2==0)ps--;
ps%=mod;
ps+=mod;
ps%=mod;
// cout<<ps<<endl;
if(n%2){
ll tt=0;
ll mo=(n+1)/2;
memset(f2.a,0,sizeof(f2.a));
f2.a[1][1]=4;
f2.a[2][1]=2;
f2.a[2][2]=1;
N=2;
f2=quick(f2,mo-1);
ll cha=f2.a[2][1];
cha%=mod;
ll muls=(qq(2,n)-1+mod)%mod;
N=3;
memset(f2.a,0,sizeof(f2.a));
f2.a[1][2]=1;
f2.a[1][1]=muls;
f2.a[3][1]=cha;
f2.a[3][3]=1;
f2=quick(f2,m-2);
ps=ps*f2.a[1][1]-f2.a[3][1];
ps=(ps%mod+mod)%mod;
cout<<ps<<endl;
}
else{
ps=ps*qq((qq(2,n)-1+mod)%mod,m-2)%mod;
cout<<ps<<endl;
}
}
return 0;
}