C语言编程.输入2个正整数a和n,求a aa aaa aa-a(n个a)之和.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 08:55:48
![C语言编程.输入2个正整数a和n,求a aa aaa aa-a(n个a)之和.](/uploads/image/f/531682-34-2.jpg?t=C%E8%AF%AD%E8%A8%80%E7%BC%96%E7%A8%8B.%E8%BE%93%E5%85%A52%E4%B8%AA%E6%AD%A3%E6%95%B4%E6%95%B0a%E5%92%8Cn%2C%E6%B1%82a+aa+aaa+aa-a%28n%E4%B8%AAa%29%E4%B9%8B%E5%92%8C.)
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>#include<math.h>int min(int x,int y)\x09\x09//求m和n的最小值{\
#includeintmain(){intm,n;intm_cup,n_cup,res;/*被除数,除数,余数*/printf("Entertwointeger:\n");scanf("%d%d",&
#include int main() { int m, n; int m_cup, n_cup,
main(){inta,b,num1,num2,temp;printf("请输入两个正整数:\n");scanf("%d,%d",&num1,&num2);if(num1
楼上是C++写的,这个是C语言版的#include#includeintmain(){intcount;int*arrayLenth;int**intArray;int*elem1,*elem2;in
#include <stdio.h> #include <string.h> #define N 200
#include#includelongfac(intn,inta){longsum;if(n==1){sum=a;}else{sum=(long)(pow(10,n-1)*a)+fac(n-1,a)
#includeintmain(){intm[10],i,j=0,k;longn;scanf("%ld",&n);k=n;while(k>0){k/=10;j++;}i=j;while(i--){m[
#include<stdio.h>int main(){ int n; do
利用辗除法公约数,再算公倍数.#include <stdio.h>void main(){ int a,&nbs
把你写的给我看一下再问:#include#includeintmain(void){inti,j,p,m,n,count;count=0;printf("Inputm:");scanf("%d",&m
#includevoidmain(){inta,b,c,max,min;printf("请分别输入a,b,c三个数:\n");scanf("%d%d%d",&a,&b,&c);max=min=a;if
#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i
/*2 (repeat=2)2 3 (a=2, n=3)8 5 (a=8, n=5)  
#include#includeintmain(void){intn,m,i,j,t;scanf("%d%d",&n,&m);i=m>n?m:n;j=m>n?n:m;while(j){t=i%j;i=
最大公约数:intGcd(inta,intb){if(a%b==0)returna;return(b,a%b);}最小公倍数:intGbs(inta,intb){returna*b/Gcd(a,b);
#includevoidmain(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=
有时间和空间要求么?简单方法如下:count=0;for(i=A;i再问:你的好像不行,这是我写的,看看怎么改一下#include#includeintmain(void){intA,B,count=
while循环for循环太多了也太乱了.我给你修改了一些,也调试成功了,你看一下吧#include#includeintmain(void){inta,b,n,count,i,x,y;scanf("%