include <stdio.h> int main(void){ int i, j, n;int p = 1, sum = 0;printf("input n = ");scanf("%d", &n);for (i=1; i<=n; i++){ p = 1;for (j=1; j<=i; j++){ p *= j;} sum += p;} printf("1!+2!+3!+...+n! = %d\n", sum);return 0...
include<stdio.h>int fun(int a){int i=1,sum=1;while(i<=a){sum*=i;i++;}return sum;}main(){int n,i=1,sum=0;printf("请输入n的值：");scanf("%d",&n);while(i<=n){sum+=fun(i);i++;}printf("结果是：%d\n",sum);}include<bits/stdc++.h>using namespace std;...
include<stdio.h> int fun(int n){ int i, sum=0;for(i=0;i<=n;i++)sum+=i;return sum;} main(){ int n;scanf("%d",&n);printf("s=%d\n",fun(n));} int
include "iostream"using namespace std;int fun(int n){ int sum=1;int i;for(i=1;i<=n;i++)sum*=i;return sum;} int main(){ int i,mark=1;int sum=0,item=0;for(i=1;i<=9;i++){ item=mark*fun(i);sum+=item;mark=-mark;} cout<<"1!-2!+3!-4!+5!-6!+7!
求1!+2!+3!...n!的结果 输入输入正整数n(1<=n<=12)输出数据计算结果 样例输入3样例输出9 188...int fun(int n)//求n!{int i=1,c=1;while(i<=n){c*=i;i++;}return c;}int main()...for(i=1;i<=a;i++){ l=i*l; s=s+l;}printf("%d\n",s);return 0;}试试~看看是不...
include <stdio.h>int fun(int n){int s=n;if(n-1) s+=fun(n-1);return s;}void main(){int n;scanf("%d",&n);printf("s=%d",fun(n));}
define N 5 int fun(int *s,int a,int n){int j;s=a;j=n;while(a!=s[j])j--;return j;} main(){int s[N+1];int k;for(k=1;k<=N;k++)s[k]=k+1;printf("%d\n",fun(s,4,N));} 首先我们应该分析出s0=4,s1=2,s2=3,s3=4,s4=5,s5=6 当j=n=5时，while（...
printf("the result is %d\n",result); return 0;}其中long factor(int n)函数就是直接求1!+2!+3!+…+n!的和 本回答由网友推荐 举报| 答案纠错 | 评论 1 0 w1s2j3229 采纳率:20% 擅长: 电脑/网络 单机游戏 电影 幽默滑稽 动漫 其他
include "stdio.h"int sum(int n);main(){ int i,n,s = 0;printf("Input n:");scanf("%d",&n);printf("%d\n",sum(n));//修改 } int sum(int n){ int i,s=0;for (i = 1; i <= n; i++){ if (i % 2)//奇数为正 s += i;else s -= i;} return s;} in...
include usingnamespacestd;intmain(){ intn;staticintsum=1;cout<<"请输入N："<<endl;cin>>n;for(inti=1;i<=n;i++){ for(intj=1;j<=i;j++){ sum=sum+j;} } cout<<"结果2为："<<sum;system("pause");return0;