代码: #include<stdio.h> int main(void) { int k, times = 0; for (int i = 1; i < 101; i++) {//将循环次数限定在[1,100] k = i; while (k != 0) { if (k % 10 == 9) {//判断k的个位上是否有9, times++; } k = k / 10;//判断k的十位是否有9,若有,则重新进入while循环 } } printf("出现%d次数字9\n", times); } 运行结果: