练习，递归求奇数的积

递归求奇数连乘的积，类似求阶乘，写出1*3*...*n的递归式，并编写出递归函数求解。

#include <iostream>using namespace std;long f(int);long y;int main(){    cout<<"输入N的值："<<endl;    int n;    cin>>n;    cout<<"N个连续奇数的积是"<<y<<endl;    return 0;}long f(int n){    long s;    if(n==1)        s=1;    else        s=f(n-1)*(2*n-1);    return s;}

上面与题目不符

#include <iostream>using namespace std;long f(int);long y;int main(){    cout<<"输入N的值："<<endl;    int n;    cin>>n;    if(n%2!=0)        y=f(n);    else        y=f(n-1);    cout<<"N以内的奇数的积是"<<y<<endl;    return 0;}long f(int n){    long s;    if(n==1)        s=1;    else        s=f(n-2)*n;    return s;}