思路:由指数型母函数的知识f(x)=(1+x/1!+x^2/2!+x^3/3!...+x^n/n!)^2+(1+x^2/2!+x^4/4!+x^6/6!...+...)^2;又由大学的泰勒公式:e^x=1+x/1!+x^2/2!+x^3/3!...+x^n/n!;e^(-x)=1-x/1!+x^2/2!-x^3/3!+...-...;所以e^x+e^(-x)=1+x^2/2!+x^4/4!+x^6/6!...+...;
所以: f(x)=e^(2x) * ((e^x+e^(-x))/2)^2
= (1/4) * e^(2x) * (e^(2x) + 2 + e^(-2x))
= (1/4) * (e^(4x) + 2*e^(2x) +1)
= (1/4) * ( (1+4x/1!+(4x)^2/2!+(4x)^3/3!+...+(4x)^n/n!) + 2*(1+2x/1!+(2x)^2/2!+(2x)^3/3!+...+(2x)^n/n!) +1)
得: x^n 项系数
a(n) = (1/4) * ((4x)^n/n! + 2*(2x)^n/n!)
= (1/4) * ( 4^n*x^n/n! + 2^(n+1)*x^n/n!)
= (4^(n-1) + 2^(n-1)) * x^n/n!
即所求 F(n) = (4^(n-1) + 2^(n-1)) % 100.
#include<cstdio> #define MOD 100 int pow_mod(int a,__int64 n) { int res=1; while(n){ if(n&1)res=res*a%MOD; a=a*a%MOD; n>>=1; } return res; } int main() { int t; unsigned __int64 n; while(~scanf("%d",&t),t){ int cas=1; while(t--){ scanf("%I64u",&n); printf("Case %d: %d ",cas++,(pow_mod(4,n-1)+pow_mod(2,n-1))%100); } printf(" "); } return 0; }