C语言编程求出最小的正整数n,使得n满足除3余2,除5余3,除7余4
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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
#includevoidmain(){intn;ints=0;printf("请输入一个正整数:");scanf("%d",&n);printf("该整数除1和其本身的因子为:\n");for(int
main(){intm,n,sum;scanf("%d",&n);for(m=1,sum=0;sum
下面的程序已经给你修改正确了:#includevoidmain(){intm,n,i,j,k;scanf("%d%d",&m,&n);for(i=m;i
#include<stdio.h>int main(){\x09int n,s=0;\x09scanf("%d",&n);\x09while
定义unsignedintn,longlongintn1,计算过程用for循环每次*10,保存到n1,最后输出n1再答:算法思路大致就是这样了,具体代码应该不难写再问:再答:哦,原来是这个再答:那么把
上一个你还没采纳呢1)intmax=0;intn;do{scanf("%d",&n);if(max再问:采纳了亲再答:第一个刚修改了第二有稍复杂点,先吃饭去。再问:谢了哈再答:回来了第二个给你补上:主
#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i
#includeintmax(intn);voidmain(){inta,b;printf("Pleaseinputnumber,endof-1:");scanf("%d",&a);while(a!=
#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=
inti,flag,M=0,N=0;\x09printf("输入素数起始范围:");\x09scanf("%d%d",&M,&N);\x09for(M;M
#includevoidmain(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=
辗转相除法求最大公约数!#includeintmain(){/*辗转相除法求x与y最大公约数*/intx,y,r,temp;x=100,y=60;if(x
main(){intp,r,n,m,temp;printf("Pleaseenter2numbersn,m:");scanf("%d,%d",&n,&m);//输入两个正整数.if(n
for(j=2;j
DimiAsIntegerFori=1To100If(iMod3=2AndiMod5=3AndiMod7=4)ThenMsgBoxiExitSubEndIfNext