c语言用递归法求Y=x+x²/2！+x³/3！+……到第n项，n和x的值由键盘输入 用递归法求:y=x+x^2/2!+x^3/3!+…………到第...
c\u8bed\u8a00\u7528\u9012\u5f52\u6cd5\u6c42y=x+x2/2!+x3/3!\u2026\u5230\u7b2cn\u9879,n\u548cx\u7684\u503c\u7531\u952e\u76d8\u8f93\u5165#include#includefloat fun(int,float,float *ans);int main(){ int n; float x,sum=0; float *ans; printf("\u8bf7\u8f93\u5165N\uff1a\n"); scanf("%d",&n); printf("\u8bf7\u8f93\u5165X(float)\uff1a\n"); scanf("%f",&x); ans=(float *)malloc((n+1)*sizeof(float)); fun(n,x,ans); for(int i=1;i!=(n+1);i++) { sum+=*(ans+i); } printf("\u6c42\u5f97\u548c\u4e3a\uff1a%f\n",sum); free(ans);}/**\u8be5\u51fd\u6570\u5c06\u6c42\u548c\u591a\u9879\u5f0f\u62c6\u5f00\uff0c\u89c2\u5bdf\u6bcf\u4e00\u9879\uff0c\u53d1\u73b0\u7b2c\uff08n+1\uff09\u9879\u7b49\u4e8e*\u7b2cn\u9879\u4e58\u4ee5\uff08x/(n+1)\uff09,\u56e0\u6b64\u53ef\u9012\u5f52\u6c42\u5f97\u6c42\u548c\u591a\u9879\u5f0f \u4e2d\u7684\u6bcf\u9879\u7684\u503c\uff0c\u7136\u540e*\u6839\u636e\u8f93\u5165N\u7684\u503c\u5f00\u8f9fN+1\u6570\u7ec4\uff0c\u6570\u7ec4\u7528\u4e8e\u5b58\u653e\u5728\u9012\u5f52\u8fc7\u7a0b\u4e2d\u4ea7\u751f\u7684\u524d\u9762\u9879\u7684\u503c\uff0c\u8fd9\u6837*\u901a\u8fc7\u4e00\u6b21\u9012\u5f52\uff0c\u4fbf\u6c42\u5f97\u6240\u6709\u9879\u7684\u503c\uff0c\u7136\u540e\u6839\u636e\u6570\u7ec4\u4e2d\u7684\u503c\u8fdb\u884c\u6c42\u548c\u5373\u53ef\u3002*/ float fun(int n,float x,float *ans){ if(n==1) { *(ans+n)=x; return x; }else{ *(ans+n)=(x/(n))*fun(n-1,x,ans); return *(ans+n); }}
#include#includefloat fun(int,float,float *ans);int main(){ int n; float x,sum=0; float *ans; printf("\u8bf7\u8f93\u5165N\uff1a\n"); scanf("%d",&n); printf("\u8bf7\u8f93\u5165X(float)\uff1a\n"); scanf("%f",&x); ans=(float *)malloc((n+1)*sizeof(float)); fun(n,x,ans); for(int i=1;i!=(n+1);i++) { sum+=*(ans+i); } printf("\u6c42\u5f97\u548c\u4e3a\uff1a%f\n",sum); free(ans);}/**\u8be5\u51fd\u6570\u5c06\u6c42\u548c\u591a\u9879\u5f0f\u62c6\u5f00\uff0c\u89c2\u5bdf\u6bcf\u4e00\u9879\uff0c\u53d1\u73b0\u7b2c\uff08n+1\uff09\u9879\u7b49\u4e8e*\u7b2cn\u9879\u4e58\u4ee5\uff08x/(n+1)\uff09,\u56e0\u6b64\u53ef\u9012\u5f52\u6c42\u5f97\u6c42\u548c\u591a\u9879\u5f0f \u4e2d\u7684\u6bcf\u9879\u7684\u503c\uff0c\u7136\u540e*\u6839\u636e\u8f93\u5165N\u7684\u503c\u5f00\u8f9fN+1\u6570\u7ec4\uff0c\u6570\u7ec4\u7528\u4e8e\u5b58\u653e\u5728\u9012\u5f52\u8fc7\u7a0b\u4e2d\u4ea7\u751f\u7684\u524d\u9762\u9879\u7684\u503c\uff0c\u8fd9\u6837*\u901a\u8fc7\u4e00\u6b21\u9012\u5f52\uff0c\u4fbf\u6c42\u5f97\u6240\u6709\u9879\u7684\u503c\uff0c\u7136\u540e\u6839\u636e\u6570\u7ec4\u4e2d\u7684\u503c\u8fdb\u884c\u6c42\u548c\u5373\u53ef\u3002*/ float fun(int n,float x,float *ans){ if(n==1) { *(ans+n)=x; return x; }else{ *(ans+n)=(x/(n))*fun(n-1,x,ans); return *(ans+n); }}
f(n) = f(n-1) + x的n次方/n的阶乘；
简化 = f(n-1) + (f(n-1) - f(n-2))*x/n;
故有以下代码(动态规划的递归解法)：
//由于这中间涉及除法，我选用了double，楼主视情况修改吧
double f[SIZE] = {0};
double getY(double x, double n)
{
if(n == 1)return x;
if(n == 2)return x*x/2 + x;
if(f[n] != -1)return f[n];
f[n] = (1+x/n)*getY(x, n-1) - (x/n)*getY(x, n-2);
return f[n];
}
#include "stdafx.h"
int n=0;
long x=0;
double sum=0;
double result=0;
double i=0;
double process(int nnn,long xxx)
{
double k=1;
double multi_sum=xxx;
if(nnn>0)
{
for(i=1;i<=nnn;i++)
{
k=k*i;
}
if(nnn==1)
{
multi_sum=xxx;
}
else
{
for(i=2;i<=nnn;i++)
{
multi_sum=multi_sum*xxx;
}
}
sum=sum+(multi_sum)/k;
process(nnn-1,xxx);
}
else
{
return sum;
}
}
int main(int argc, char* argv[])
{
printf("请输入X的值：\n");
scanf("%d",&x);
printf("请输入n的值：\n");
scanf("%ld",&n);
process(n,x);//result=process(n,x);会显示乱码，我也不知道为什么所以直接用SUM做全局变量，然后赋值
result=sum;
printf("结果：%lf\n",result);
return 0;
}
下面是公式，如果这样还不会写，说明笨到家了，呵呵。
f(n) = f(n-1) * n n>0; n==0 f(0)=1
g(n) = g(n-1) + x^n/f(n)
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