C语言编写程序:输入两个正数m,n输出mn间所有的fibonacci数
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#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i
#includevoidmain(){inta,b;printf("Inputtwointegers:");scanf("%d%d",&a,&b);printf("和:a+b=%d\n",a+b);p
楼上两位的代码可能有点小问题s += m*m+1/m;改成s += m*m+1.0/m;再问:您好,请您帮忙写一个完整的编程好么?谢谢了再答:#include&
intiX,iY,iZ;scanf(.)if(iX>iY)iZ=iX-iY;elseiZ=iY-iX;
再简化一下#include#defineN10voidmain(){inta[N],i,s1=0,s2=0;for(i=0;i
#include"stdio.h"main(){intm,n,t,h,a,b,q;printf("输入m,n的值:");scanf("%d%d",&m,&n);a=m;b=n;if(n>m){t=m;
首先明确gbs(最小公倍数)=m*n/gys(最大公约数);然后求最大公约数用欧几里得辗转相除法;代码很短的.#include"stdio.h"intgys(intm,intn){returnn==0
1:#includeintmain(){intnum,sum=0;while(scanf("%d",&num)==1&&num>0){sum+=num;\x09}printf("结果为%d",sum)
#include <stdio.h>int main() { int m, n; int m_cup,&nb
输入两个整数,中间空格:#include<stdio.h>main(){\x09int a,b;\x09printf("Input a,b:\n")
#include#incluedvoidmain(){inta;intb;intc,d;a=1500,b=350;c=a/b;//商数d=a%b;//余数pritf("商数为:%d\n",c);pri
#includeintmain(){inta,b,c,d=0;scanf("%d%d",&a,&b);c=a/b;d=a%b;printf("商是:%d\n",c);printf("余数是:%d\n"
#includeintmain(){intx,y,a;scanf("%d",&x);scanf("%d",&y);a=x*y;printf("Theaverageis%d\n",a);return0;
10个数正数和:#include<stdio.h> #define N 10 void main() {
#include#includemain(){\x09longintx;\x09inti=0;printf("input:");\x09scanf("%ld",&x);do{x=x/10;i++;\x
#includevoidmain(){\x09intm,n,k,min,max,i;\x09scanf("%d%d",&m,&n);\x09if(n0;i++)\x09{\x09\x09if((m%i
#include"stdio.h"intmain(){inta,b;scanf("%d%d",&a,&b);if(b==0)b=1;printf("和%d\n",a+b);printf("积
#include"stdio.h"voidmain(){inta,b,c,d;printf("请输入两个正整数:");scanf("%d%d",&a,&b);if(a==b)printf("您输入两数
#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min
#include <stdio.h>int abc(int x,int y);void main(){int n1,n2,i;