C语言输入正整数n,求数列 s=1 2 3 ... n ,s的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/14 23:23:56
#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i
main(){inta,b,num1,num2,temp;printf("请输入两个正整数:\n");scanf("%d,%d",&num1,&num2);if(num1
#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for
#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;
#include<stdio.h>int main(){\x09int n,s=0;\x09scanf("%d",&n);\x09while
#include <stdio.h>int main() { int m, n; int m_cup,&nb
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
#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;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(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=
使用了数组的动态分配,看不懂的话我还有其他方案#include#includeboolprime(intnum){boolflag=true;if(num
#include <stdio.h>#include <string.h>main(){\x05int n=0;\x05int s=0;
#include"stdio.h"main(){intm,n,i;longintsum=1,sum1=1,sum2=1,t;printf("请输入m>n\n");scanf("%d",&m);scan
#includemain(){intn,sum=0;scanf("%d",&n);for(inti=0;i
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
#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min
#includeintmain(void){intn=0,m=0,i=0,j=0,k=0;scanf("%d",&n);while(n--){\x09scanf("%d",&m);\x09for(j=
#includevoidmain(){intn,i=1;doublea=0,x;printf("请输入要求X的前几项之和:");scanf("%d",&n);printf("请输入x的前m项(m>=n