题解 | #奇偶校验#

#include <stdio.h>
#include <string.h>
#include<stdlib.h>

int main() {
    char a[100];
    while (scanf("%s", &a) != EOF) { // 注意 while 处理多个 case
        // 64 位输出请用 printf("%lld") to 
        //printf("%s:%d\n", &a,strlen(&a));
        int i,j;
        //校验输入的合法性
        for(i=0;i<strlen(&a);i++){
            if((a[i]>='0'&&a[i]<='9')||(a[i]>='a'&&a[i]<='z')||(a[i]>='A'&&a[i]<='Z')){
                int x=a[i],m,count=0;
                char outString[9]={'0'};
                for(m=8-1;m>=0;m--){
                    if(x%2==1) count++;
                    outString[m]=(x%2+48);
                    x=x/2;
                }
                if(count%2==0) outString[0]='1';
                outString[8]='\0';
                printf("%s\n",&outString);
                continue;
            }
            printf("输入不合法\n");
        }

    }
    return 0;
}