c语言一个正整数与3的和是5
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#includemain(){inti=0,n=5;while(n){if(((i+3)%5==0)&&((i-3)%6==0)){\x05\x05printf("%d",i);\x05\x05n--
可参考如下scanf("%d",&num);printf("%d\n",num/100);printf("%d\n",num%10);---------------------------------
#include"stdio.h"#include"math.h"intjiecheng(inti){\x09intk=1;\x09while(i>=1)\x09{\x09\x09k*=i;\x09\
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;
#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"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){inti,j,k;for(i=15;i
#includeintmain(){printf("请输入一个数:");inta;scanf("%d",&a);printf("\n%d=",a);for(inti=2;i
#include <stdio.h>void main (){ int a[10]; int i,k=0,n,min,
#includeintmat[10][10];voidmain(){intn,i,j;intok=1;scanf("%d",&n);for(i=0;i
#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;
#includeintsign(intx);intmain(void){intx,y;intrepeat,ri;/*repeat:1至10的整数*/scanf("%d",&repeat);for(ri
#includefloatx;intn,i;floats=1.0;voidmain(){printf("PleaseInputx:");scanf("%f",&x);printf("PleaseInp
这种判断语句就是程序上的1+1命令,有必要在这里问吗,而且你又问错地方.再问:我把什么地方想错了啊????我感觉我现在像是钻进了个牛角尖出不来了再答:做个递增单循环,当递增变量大于判断的数时中断,判断
#includeintgcd(intm,intn)//最大公约数{intt;if(m再问:t=n,n=m,m=t;if(n==0)returnm;elsereturngcd(n,m%n);求解释。还有
main(){intp,r,n,m,temp;printf("Pleaseenter2numbersn,m:");scanf("%d,%d",&n,&m);//输入两个正整数.if(n
与13的和为5的倍数的正整数有2,7,12,…,2+5×n,…(n为正整数),与13的差为3的倍数的正整数有1,4,7,…,1+3×n,….所以这个正整数最小是7.故答案为:7.
#includeintmain(void){intn=0,m=0,i=0,j=0,k=0;scanf("%d",&n);while(n--){\x09scanf("%d",&m);\x09for(j=
#includevoidmain(){voidshusu_(intz);intn;printf("pleaseentern:\n");scanf("%d",&n);shusu_(n);}voidshu
#includevoidmain(){longintx,ge,shi,bai,qian,wan,bit;scanf("%ld",&x);ge=0;shi=0;bai=0;qian=0;wan=0;bi
#include#includeintmain(void){charsx[80],longest[80];inti,n;intrepeat,ri;scanf("%d",&repeat);for(ri=