题解 | #二进制数#
二进制数
https://www.nowcoder.com/practice/103dd589fed14457a673c613d8de3841
#include <stdio.h> void fun(int n); int main() { int a, b; while (scanf("%d", &a) != EOF) { // 注意 while 处理多个 case // 64 位输出请用 printf("%lld") to fun(a); } return 0; } void fun(int n) { int x, t = n, cnt = 0; while (t > 0) { x = t % 2; t /= 2; cnt++; } int s[cnt]; int i = 0; t = n; while (t > 0) { s[i++] = t % 2; t /= 2; } for (i = 0; i < cnt; i++) { printf("%d", s[cnt - i - 1]); } printf("\n"); }