c语言计算输入一个数计算1 2 ... n的值

#includeintGetNumber(intn)//用递归来实现很简单{intsum=0;if(n/10!=0){\x09sum+=GetNumber(n/10);}sum+=n%10;retur

求和应当是：#include#includemain(){inti,up=0,down=0,data[10];for(i=0,i0)up+=data[i];elsedown+=data[i];}pri

用数组main(){inti,m=0,n=0,a[10];printf("input10numbers:\n");for(i=0;i

#include"stdio.h"intmain(){inta[10][10]={0};intn,i,j,sum1=0,sum2=0;scanf("%d",&n);for(i=0;i

#includeintmain(){inta,sum=0;scanf("%d",&a);while(a){sum+=a%10;a/=10;}printf("sum=%d\n",sum);return0

已经运行确认了：#includemain(){intavg,i,sum=0,a[11];printf("请输入十个整数\n");for(i=1;i

#includeusingnamespacestd;main(){doublex,y;cout

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

代码：#include <stdio.h>int sum(int n){\x05int s = 0;\x05while(n>

#include"stdio.h"intmain(){inta,i,sum=0;printf("请输入3位的整数\n");scanf("%d",&a);for(i=0;i

#includeintmain(){inti;intj;intk;inta[4][4];for(i=0,k=1;i{intsumrow=0;for(j=0;j{a[i][

#include"stdio.h"intmain(){    inta[10][10];    intm,n,tmp,i

#includevoidmain(){inta;printf("请输入一个三位数：");scanf("%d",&a);printf("各位数上的乘积为：%d\n",(a/100)*(a%10)*(a/

/*Note:YourchoiceisCIDE*/#include"stdio.h"#include"math.h"#definePI3.1415926//把;去掉voidmain(){//#defi

#includemain(){floata,b,s;printf("pleaceinputa,b:\n");scanf("%f%f\n",&a,&b);s=a+b;printf("%f\n",s);/

#include#include#defineE2.7182voidmain(void){floatx;floaty;scanf("%d",&x);if(x>=-1)y=pow(x,2)+1;

要是是整形的话可以这样做：inta=2134;intcount=0;while(a!=0){count++;a/=10;}printf("%d",count);以上可以借鉴下,你自己思考下这个数是浮点

当然不行了,^是C语言中的异或运算符.计算平方可以用b=a*a

#includeintmain(){intlength,width,perimeter;printf("长：");scanf("%d",&length);printf("宽：");scanf("%d"

(int)(x+y)=10a%3*10%2/4=2*10%2/4=0x+0=3.5%求余