题解 | #鸡兔同笼#
鸡兔同笼
https://www.nowcoder.com/practice/fda725b4d9a14010bb145272cababef1
#include <stdio.h> #include <limits.h> #include <stdlib.h> #include <math.h>//注意famx头文件 int main() { int a; while (scanf("%d", &a) != EOF) { int min = INT_MAX; int max = INT_MIN; for (int i = 0; i <= a / 2; i++) { for (int j = 0; j <= a / 4; j++) { if (i * 2 + j * 4 == a) {//满足条件时更新 max = fmax(max, i + j); min = fmin(min, i + j); } } } if(min==INT_MAX) printf("0 ");//注意不存在的情况 else printf("%d ",min); if(max==INT_MIN) printf("0 \n"); else printf("%d \n",max); } return 0; }