问题：啤酒2块1瓶，4个盖换1瓶，2个空瓶换1瓶，10块可以喝几瓶
#include<stdio.h>
#define PRICE 2
#define BOTTLE_PEICE 2
#define BOTTLECAP_PRICE 4
int main() {
int money = 4; //钱的数目
int bottle = 0; //当前的瓶子数
int bottleCap = 0; //当前的瓶盖数
int count = 0; //喝酒的瓶数
while (money >= PRICE || bottle >= BOTTLE_PEICE || bottleGap >= BOTTLECAP_PRICE) {
while (money >= PRICE) {
money -= PRICE;
bottle++;
bottleCap++;
count++;
}
while (bottle >= BOTTLE_PEICE) {
bottle -= BOTTLE_PEICE;
bottle++;
bottleCap++;
count++;
}
while (bottleGap >= BOTTLECAP_PRICE) {
bottleGap -= BOTTLECAP_PRICE;
bottle++;
bottleCap++;
count++;
}
}
printf("可以喝 %d 瓶\n",count);
printf("剩余money：%d , 剩余空瓶：%d ,剩余瓶盖：%d\n",money,bottle,bottleCap);
return 0;
}
