如何用递归的方法求1!+2!+......+n!? 使用递归方法,编写一个求解S=1+2+3+…+n的函数
C\u8bed\u8a00\u5982\u4f55\u7528\u9012\u5f52\u7b97\u6cd5\u6c421!+2!+3!+...n!#include
float fun(int n)
{
if(n==1) return 1;//\u5982\u679cn=1\u5219\u76f4\u63a5\u8fd4\u56de1
return n*fun(n-1);//\u5426\u5219\u8fd4\u56den*fun(n-1),\u4ee5\u6b64\u8ba1\u7b97n\u7684\u9636\u4e58\uff0c\u8fd9\u6761\u8bed\u53e5\u5c31\u662f\u9012\u5f52\u4f53
}
void main()
{
int i;
float sum=0;
for(i=1;i<=n;i++){
sum+=fun(i); //\u5faa\u73af\u8c03\u7528\uff0c\u7528sum\u7d2f\u8ba1
}
printf("sum=%.2f\n",sum);
}
#include "stdio.h"
int fun(int num)
{int sum;
if( num==1) sum=1;
else sum=num+fun(num-1);
return sum;}
void main()
{ int n,s;
printf("n=");
scanf("%d",&n);
s=fun(n);
printf("s=%d\n",s);
}
//阶乘递归
public static int jieChengNew(int y){
int result2=y;
if(y==0||y==1)
return 1;
else{
result2=result2*jieChengNew(y-1);
return result2;
}
}
//求和递归
public static int addNum(int x){
int result=x;
if(x==0||x==1)
return 1;
else{
result=jieChengNew(x)+addNum(x-1);
return result;
}
}
public static void main(String []args){
System.out.println("小于等于1所有阶乘之和为："+addNum(1));
System.out.println("小于等于2所有阶乘之和为："+addNum(2));
System.out.println("小于等于3所有阶乘之和为："+addNum(3));
System.out.println("小于等于5所有阶乘之和为："+addNum(5));
System.out.println("小于等于10所有阶乘之和为："+addNum(10));
}
}
我这个是用迭代完成的,你看下吧.
public class Diedai
{
public static void main(String [] args)
{
Diedai d = new Diedai();
//直接调用funAdd()方法即可
int num = d.funAdd(4);
System.out.println(num);
}
//阶乘方法
public int funMul(int num)
{
int count = 1 ;
if(num >= 1)
{
count = num * funMul(num - 1 ) ;
}
return count ;
}
//循环加
public int funAdd(int num)
{
int count = 0 ;
if(num >= 1)
{
count = funMul(num) + funAdd(num - 1) ;
}
return count ;
}
}
public class jiechen{
public static int jc (int x)
{
if (x < 0) {
throw new IllegalArgumentException("x must be>=0");
}
if (x <= 1) {
return 1;
} else
return x * jc (x - 1);
}
public class sum()
{
int sum=0;{
for(i=1;i<=x;i++)
sum=sum+jc(x);
}
public static void main(String args[]) {
System.out.print(sum(2));
}
}
public class Demo{
public static void main(String [] args){
Demo demo = new Demo();
System.out.println(demo.getResult(3));
}
public int jieCheng(int number){
if(number < 1){
return -1;
}
if(number == 1){
return 1;
}else{
return number * jieCheng(number-1);
}
}
public int getResult(int number){
int result = 0;
for(int i = 1; i <= number; i++){
result += jieCheng(i);
}
return result;
}
}
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绛旓細//闃朵箻閫掑綊 public static int jieChengNew(int y){ int result2=y;if(y==0||y==1)return 1;else{ result2=result2*jieChengNew(y-1);return result2;} } //姹傚拰閫掑綊 public static int addNum(int x){ int result=x;if(x==0||x==1)return 1;else{ result=jieChengNew(x)+ad...
绛旓細int n){ if(n==1) return 1;//濡傛灉n=1鍒欑洿鎺ヨ繑鍥1 return n*fun(n-1);//鍚﹀垯杩斿洖n*fun(n-1),浠ユ璁＄畻n鐨勯樁涔橈紝杩欐潯璇彞灏辨槸閫掑綊浣 } void main(){ int i;float sum=0;for(i=1;i<=n;i++){ sum+=fun(i); //寰幆璋冪敤锛岀敤sum绱 } printf("sum=%.2f\n",sum);} ...
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绛旓細//1鍒皀鍜岀殑閫掑綊鏂规硶 public static int add(int max){ int sum=0;if(max>0){ sum=max+add(max-1);} return sum;} //1鍒皀骞虫柟鍜岀殑閫掑綊鏂规硶 public static int squr(int max){ int sum=0;if(max>0){ sum=max*max+squr(max-1);} return sum;} //鏁扮粍鍏冪礌涔嬪拰 public static ...
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绛旓細int i;if(n == 1)return 1;else { i = fun(n-1);return n*i;} }long sum(int n){ int i,j;if(n == 1)return 1;else { i = fun(n);j = sum(n-1);return i+j;} } int main(){ int n;long s;scanf("%d",&n);s = sum(n);printf("%ld\n",s);return 0;...
绛旓細include "stdio.h"int sum(int n){if(n>1)return n+sum(n-1);elsereturn 1;}int main(){int n;scanf("%d",&n);printf("%d\n",sum(n));}
