输入两个正整数a和n,求a aa aaa - aa-a(n个a)之和.

main(){inta,b,num1,num2,temp;printf("请输入两个正整数:\n");scanf("%d,%d",&num1,&num2);if(num1

在VS2010上测试通过：#includeusingnamespacestd;boolis_prime(intx){\x09inttmp=x/2;\x09for(inti=2;i>n;\x09for(

#include#includelongfac(intn,inta){longsum;if(n==1){sum=a;}else{sum=(long)(pow(10,n-1)*a)+fac(n-1,a)

下面用到了递归解决,不知楼主能否看懂.不懂用百度hi和我私聊我也很乐意.递归只是求最大公约数,通过最大公约数求最小公倍数.#include"stdio.h"voidmain(){intm,n,d,e;

#include <stdio.h>int main() { int m, n; int m_cup,&nb

#include#includeintmain(void){intm,n,r;ints;printf("输入两数：");scanf("%d%d",&m,&n);s=m*n;while(n!=0){r=

先辗转相除法求最大公约数,再将两数之积除以最大公约数,即得到最小公倍数#includeintgetGCD(inta,intb){intr;while((r=a%b)!=0){a=b;b=r;}retu

#include <stdio.h>int isPrimeNum(int x)//判断是否为素数 {    

ints=0;for(inti=0;i

#includeintmain(){intm,n,temp,p,r;scanf("%d%d",&m,&n);if(n

/*2  (repeat=2)2 3 (a=2, n=3)8 5 (a=8, n=5)   

输入两个正整数m和n,求其最大公约数和最小公倍数.用辗转相除法求最大公约数算法描述:m对n求余为a,若a不等于0则m0){m_cup=m;n_cup=n;res=m_cup%n_cup;while(r

#include"stdio.h"intis(intnumber){inttemp=number,sum=0;if(temp0){sum+=(temp%10)*(temp%10)*(temp%10);

#includevoidmov(int*x,intn,intm);intmain(void){inti,m,n;inta[80];scanf("%d%d",&n,&m);for(i=0;iscanf(

voidmain(){intm,n,i,t;intfactorsum(intnumber);//声明一个方法factorsum（intnumber）printf("Inputm(m>=1):")

#include"iostream"usingnamespacestd;boolis(intnumber){intsum=0,num=number;while(num>0){sum+=(num%10)

#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min

占天时地利人和取九州四海财宝横批：财源不断

C语言版：includeincludevoidmain(){inta,n,S;inti;printf("a=?\n")scanf("%d",&a);printf("\nn=?\n")scanf("%d

#include#defineMax90intmain(){longlongf[Max];inti,m,n;f[0]=1;f[1]=1;for(i=2;i