摘要：递推公式：E(t)=(1+t+t^2/2!+......+..)^2*(1+t^2/2!+t^4/4!+..+)^2 =e^2t*((e^t+e^-t)/2)^2 =1/4(e^4t+2*e^2t+1) =sigma(1/4*[4^n+2*2^n+1]*t^n/n!) n=0,1,2,，， ==> a(n)=1/4(4^n+2*2^n)/* * poj3734.c * * Created on: 2011-10-11 * Author: bjfuwangzhu */#include<stdio.h>#define nmod 10007int modular_exp(int a 阅读全文

摘要：/* * hdu1521.c * * Created on: 2011-8-31 * Author: 王竹 */#include<stdio.h>#include<string.h>#define nmax 15int num[nmax];double res[nmax], fac[nmax], temp[nmax];void init(int n) { int i; fac[0] = 1; for (i = 1; i <= n; i++) { fac[i] = fac[i - 1] * i; }}int main() {#ifndef ONL... 阅读全文

摘要：由题知：(1+x/1!+x^2/2!+``+x^n/n!)^2*(1+x^2/2!+```)^2由e^x=1+x/1!+x^2/2!+```知原式=e^(2*x)*((e^x+e^(-x))/2)^2 =(1/4)*(e^(2*x)+1)^2 =(1/4)*(e^(4*x)+2*e^(2*x)+1) =(1/4)*(sia(4^n)*(x^n/n!)+2*sia(2^n)*(x^n/n!)+1) 由以上式子可知： x^n/n!的系数为(4^n+2*2^n+1)/4=4^(n-1)+2^(n-1)+1/4对于本题只需要计算(4^(n-1)+2^(n-1))%100即可。其中设计大数取余... 阅读全文