利用c语言输出一个数的百位数字

intmain(){\x09intnum=456;\x09intg,s,b;\x09g=num%10;//个位\x09s=num%100/10;//十位\x09b=num/100;//百位\x09pr

可参考如下scanf("%d",&num);printf("%d\n",num/100);printf("%d\n",num%10);---------------------------------

1#include"stdio.h"main(){intn,i;scanf("%d",&n);printf("%d",n/100+n%100/10+n%10);system("PAUSE");}2#i

#includeintmain(void){inta;intge,shi,bai;scanf("%d",&a);ge=a%10;shi=a/10%10;bai=a/100;printf("%d\n",

用%（求余）.比如你输入的是321321%100=21用321-(321%100)再除以100即可得：（321-(321%100)）100=3同样,用21%10=1再除以10即可得：（（321%100

inta,b;scanf("%d%d",&a,&b);if(a>100||b>100){if(a>100)printf("%d\n",a);if(b>100)printf("%d\n",b);}els

voidmain(){doublea,b,c;scanf("%lf%lf",&a,&b);c=a*a+b*b;if(c>100)printf("a^2+b^2=%lf\n",c);elseprintf

#include#includeusingnamespacestd;doublemin(double*x,double*y);intmain(){doublea,b;cin>>a;cin>>b;if(

#includeintmain(){intnum;inta,b,c;scanf("%d",&num);a=num/100;b=(num%100)/10;c=num%10;printf("a=%d,b=

#includeusingnamespacestd;voidmain(){inta,b,c;coutb;c=a+b;cout

太简单了吧#includevoidmain(){intA,bai,shi,ge;printf("请输入一个三位整数A：");scanf("%d",&A);bai=A/100;shi=A%100/10;

只是三位数的话,下面的代码就可以了!#includeintmain(){intn;inta,b,c;scanf("%d",&n);a=n%10;b=n/10%10;c=n/100;printf("%d

11行printf("%02d",n);换行printf("%02d\n",n);19行if(n=t),双等号,22行分号去掉.

intmain(intargc,char*argv[]){inti=0,j,tmp;intarr[4]={0};scanf("%d",&j);printf("yourinputwas:%d\n\n",

//楼主,这里的大于100,指的是它们的和大于100,还是各个数字呀?//下面的例子,理解为各个数字大于100,如果理解有误,楼主可以做简的修改#includevoidmain(){inta,b,c,

数字矩形?n=3,输入1,2,3；n=4输入1,2,3,4那n=k,就是输出1,2,3…………k?那就是for(i=1;i再问：其实我是打错了。。。我想问的是可变数字矩阵。。。

sum=0;while(a!=0){sum=sum+a%10;a=a/10;}

#includeintfindmax(inta,intb){if(a>b)returna;elsereturnb;}intmain(){inta,b,c;scanf("%d%d%d",

#include#defineN100intmain(){intn,a[N],i,m,sum=0;scanf("%d",&n);for(i=0;i

#includeusingnamespacestd;main(){inti,j;for(i=1;i