c语言输入m,n,其中n表示有n个人围成一圈,顺序排号,从第一个人循环报数

#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for

一次循环就解决的,不是很难,为什么不自己尝试一下呢?给你提示一下,却是不会再给你说吧!真正想学习,最好多练习一下好不好?累乘初始值为1,循环变量循环n次,每次累乘m,循环结束即为m的n次方的值

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#includeintmain(){intM,N;int*a;inti,s=0,n=1;printf("M:");scanf("%d",&M);printf("N:");scanf("

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(){intm,n,sum,i;sum=0;scnaf("%d%d",&m,&n);for(i=m;i再问：计算n!怎么写啊再答：#includeintmain(){i

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

#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

/*mn=1001000100121144225400441484676900Pressanykeytocontinue*/#include <stdio.h>#include&

POW(a,b)表示a的B次方头文件是#include"math.h"

#includelongfun(longa);voidmain(){longm,n;printf("pleaseinputthe'm'\n");scanf("%ld",&m);printf("plea

程序清单：#include#include#include#defineMAX_SIZE20//最大单词长度//定义链表节点typedefstructLinkNode{charword[MAX_SIZ

#include<stdio.h>#include<stdlib.h>#include<string.h>typedef struct Link

if(m>=6){count=0;for(number=m;numberk是为了预防i==k且都是素数的情况){judge=false;//有一个不是素数就不行break;}}if(judge)//如

#includeintmain(void){intn=0,m=0,i=0,j=0,k=0;scanf("%d",&n);while(n--){\x09scanf("%d",&m);\x09for(j=

for(j=2;j

#includeintmain(void){intn,i,j,k,x[100],min,p,t;while(scanf("%d",&n)!=EOF&&n!=0){for(i=0;i

inti,a,n;int*pl=NULL;printf("inputn:");scanf("%d",&n);if(n>0){pl=(int*)malloc(n*sizeof(int));}for(i=

#include#defineM256#defineN256voidmain(){inta[M][N],i,j;printf("Inputdata:\n");for(i=0;i