十进制数字1--255之间
转换8位2进制数字 满足 3位=1 && 7位=1  的数有多少个

楼上的 && 楼上的 &&。。。。。。都不合我的意图

我自己研究吧

int main(void)
{

    for(int i=1;i<256;i++)
    {
        if((i&68)==68) printf("%4d",i);
    }
    printf("\n");
    return 0;
}

(10) is   68 (2) is 1000100
(10) is   69 (2) is 1000101
(10) is   70 (2) is 1000110
(10) is   71 (2) is 1000111
(10) is   76 (2) is 1001100
(10) is   77 (2) is 1001101
(10) is   78 (2) is 1001110
(10) is   79 (2) is 1001111
(10) is   84 (2) is 1010100
(10) is   85 (2) is 1010101
(10) is   86 (2) is 1010110
(10) is   87 (2) is 1010111
(10) is   92 (2) is 1011100
(10) is   93 (2) is 1011101
(10) is   94 (2) is 1011110
(10) is   95 (2) is 1011111
(10) is  100 (2) is 1100100
(10) is  101 (2) is 1100101
(10) is  102 (2) is 1100110
(10) is  103 (2) is 1100111
(10) is  108 (2) is 1101100
(10) is  109 (2) is 1101101
(10) is  110 (2) is 1101110
(10) is  111 (2) is 1101111
(10) is  116 (2) is 1110100
(10) is  117 (2) is 1110101
(10) is  118 (2) is 1110110
(10) is  119 (2) is 1110111
(10) is  124 (2) is 1111100
(10) is  125 (2) is 1111101
(10) is  126 (2) is 1111110
(10) is  127 (2) is 1111111
(10) is  196 (2) is 11000100
(10) is  197 (2) is 11000101
(10) is  198 (2) is 11000110
(10) is  199 (2) is 11000111
(10) is  204 (2) is 11001100
(10) is  205 (2) is 11001101
(10) is  206 (2) is 11001110
(10) is  207 (2) is 11001111
(10) is  212 (2) is 11010100
(10) is  213 (2) is 11010101
(10) is  214 (2) is 11010110
(10) is  215 (2) is 11010111
(10) is  220 (2) is 11011100
(10) is  221 (2) is 11011101
(10) is  222 (2) is 11011110
(10) is  223 (2) is 11011111
(10) is  228 (2) is 11100100
(10) is  229 (2) is 11100101
(10) is  230 (2) is 11100110
(10) is  231 (2) is 11100111
(10) is  236 (2) is 11101100
(10) is  237 (2) is 11101101
(10) is  238 (2) is 11101110
(10) is  239 (2) is 11101111
(10) is  244 (2) is 11110100
(10) is  245 (2) is 11110101
(10) is  246 (2) is 11110110
(10) is  247 (2) is 11110111
(10) is  252 (2) is 11111100
(10) is  253 (2) is 11111101
(10) is  254 (2) is 11111110
(10) is  255 (2) is 11111111

#include<stdio.h>

int i=0;
void dec2yyy(int yyy,int source,char* bin)
{
    if(source==0) return;
    dec2yyy(yyy,source/yyy,bin);
    if((source%yyy)<10) bin[i]=(char)(source%yyy+0x30);
    else bin[i]=(char)(source%yyy+0x37);
    i++;
}

int main(void)
{
    char dest[10]={'\0'};
    for(int j=1;j<256;j++)
    {
        if((j&68)==68)
        {
            dec2yyy(2,j,&dest[0]);
            i=0;
            printf("(10) is %4d (2) is %s\n",j,&dest[0]);
        }
    }
    printf("\n");
    return 0;
}