Problem: "red virus" problem (generating function)

topic

Portal

Thinking

Generating function directly up hate

We first consider selecting these letters, the only option was considered different letters different schemes

For \ (A \) and \ (C \)

\(G(x)=\sum_{i=0}^{\infty}\frac{1}{(2i)!}x^{2i}\)

For \ (B \) and \ (D \)

\(F(x)=\sum_{i=0}^{\infty}\frac{1}{i!}x^i\)

After written, these two forms is a routine of things

\(G(x)=\frac{e^x+e^{-x}}{2}\\F(x)=e^x\)

They will roll up

\(\begin{aligned}H(x)&=G^2(x)F^2(x)\\&=(\frac{e^x+e^{-x}}{2})^2*e^{2x}\\&=\frac{e^{4x}+2e^{2x}+1}{4}\\&=\frac{\sum_{i=0}^{\infty}\frac{1}{i!}(4x)^i+2*\sum_{i=0}^{\infty}\frac{1}{i!}(2x)^i+1}{4}\end{aligned}\)

We just need \ (x ^ n \) coefficient, do not forget our definition of the generating function, but also need to look at the whole arrangement

\ (Years = n! \ Frac {\ frac {4 ^ n} {n!} + 2 * \ frac {2 ^ n} {n!}} {4} = \ frac {4 ^ n + 2 * 2 ^ n} {4} = 4 ^ {n-1} + 2 ^ {n-1} \)

Code

#include<iostream>
using namespace std;
const int mod=100;
int qkpow(int a,long long b)
{
    if(b==0)
        return 1;
    if(b==1)
        return a;
    int t=qkpow(a,b/2);
    t=(t*t)%mod;
    if(b%2==1)
        t=(t*a)%mod;
    return t;
}
int t;
long long n;
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>t)
    {
        if(!t)
            break;
        for(int i=1;i<=t;i++)
        {
            cin>>n;
            cout<<"Case "<<i<<": "<<(qkpow(4,n-1)+qkpow(2,n-1))%mod<<'\n';
        }
        cout<<'\n';
    }
    return 0;
}