C语言输入一个正整数,输出它的所有质数因子.
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![C语言输入一个正整数,输出它的所有质数因子.](/uploads/image/f/532336-40-6.jpg?t=C%E8%AF%AD%E8%A8%80%E8%BE%93%E5%85%A5%E4%B8%80%E4%B8%AA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E8%BE%93%E5%87%BA%E5%AE%83%E7%9A%84%E6%89%80%E6%9C%89%E8%B4%A8%E6%95%B0%E5%9B%A0%E5%AD%90.)
voidmain(){intn=-1;while(n
因为要求从高位输出,第一个循环求出输入得数多少位,并用power保存相关信息(如:输入1,power=1,输入123,power=100,输入1234,power=10000,输入12345,powe
intmain(intargc,char*argv[]){intinput;intret,num;printf("Inputaint:");scanf("%d",&input);for(;;){num
很简单,用while#includevoidmain(){intn;intm;printf("输入一个数:");scanf("%d",&n);while(n>=1){m=n%10;
#includevoidmain(){\x09intn,i,j,t,a[10000];\x09printf("请输入正整数N:");\x09scanf("%d",&n);\x09printf("请依次
main(){longinta;intb,c,d,e,f;scanf("%ld",&a);while(a>=1&&a=1&&a/10000=1&&a/1000=1&&a/100=1&&a/10=1&&
#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}
#includemain(){intn,c=0;printf("输入一个小于等于一万的整数:");scanf("%d",&n);if(n>10000&&n
#includeintmain(){intm[10],i,j=0,k;longn;scanf("%ld",&n);k=n;while(k>0){k/=10;j++;}i=j;while(i--){m[
只学循环完全可以办到,关键是要动脑啊#includevoidmain(){\x09intn,m,max,i;\x09max=0;\x09printf("PleaseInputANumber:\n");
charch;scanf("%c",&ch);printf("%c\n",ch+1);
只是三位数的话,下面的代码就可以了!#includeintmain(){intn;inta,b,c;scanf("%d",&n);a=n%10;b=n/10%10;c=n/100;printf("%d
#includevoidmain(){intn,m=1,i,j;printf("请输入一个正整数:");scanf("%d",&n);i=n,j=n;while(i>0){m*=i;
方法一://用数学函数#include#includevoidmain(){inta;scanf("%d",&a);printf("%d\n",abs(a));}方法二://判断#includevoi
#include#includeintmain(void){\x09chara[20];\x09printf("请输入数字");\x09scanf("%s",a);\x09printf("%c",a[
x%10=a1,x/10%10=a2,x/100=a3,a=a1*100+a2*10+a3再答:判断是几位:if(a>100)3if(a>1000)4if(a>10000)5再问:再问:能帮我看一下错
intprime(intx){\x09inti;\x09for(i=2;i
#includemain(){inti,j,N,t,k=0;intscore[100];printf("请输入整数的个数N:\n");scanf("%d",&N);printf("请输入N个整数:\n
#includemain(){intn,sum=0;scanf("%d",&n);for(inti=0;i
修改如下:1.主函数中intcountdigit,digit=2,number中,countdigit重复定义,改为其他变量,比如x2.改为scanf("%d",&number);3.printf("