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用c语言解决输入一个八位数,求由后四位数字组成的...

#include int main(){ char s[9]; int a[4]; int i,j,t; scanf("%s",s); for(i = 4;i

#include#includevoid c(void){ char ch; while((ch=getchar())!='\n'&&ch!=EOF);}int main(){ int num; while(1) { while(1) { printf("input a number of 8 bit:"); scanf("%d",&num); c(); if(num99999999) { printf("input err!\n"); contin...

#includeint main(){ int num = 12345678, a[8], i = 7; while (num > 0) { a[i--] = num % 10; num /= 10; } for (i = 0; i < 8; i++) printf("%d", a[i]); printf("\n"); return 0;}

很简单的事情,一位16进制代表4位二进制,所以你只要把32位数转化成4个2位的16进制数就可以了,当然首先要先把32位的数转看成16进制表示,不需要你去转化,因为系统对数据的处理,这里我们以简单的整型数据举例,比如0xAAFCE8C9,你要转化成4个1...

#include int countNumber(int number, int countNumber) { //统计number中有多少个countNumber int count=0; if (number0) { if (number%10==countNumber) { count++; } number/=10; } return count;}int main() { int number; scanf("%d",&num...

#include

可以通过位域来实现. 例如: struct bs { int a:8; };

unsigned short a = 0xA4F5; unsigned char b = (a & 0xFF00)>>8; unsigned char c = a & 0x00FF; unsigned short d = (b

int check(int a,int b,int c){ int test[9]={0},i; long num = a*1000000+b*1000+c; if(b>1000 || c>1000)return 0;//确保是三位数 for(i = 0;i < 10;i++) { int temp = num%10; num/=10; temp--; if(test[temp] == 0) { test[temp] = 1; } els...

#include main() { int num; printf("请输入一个数据 "); scanf("%d",&num); printf("%08d\n",num); }

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