喝汽水问题

//第一次尝试：
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
//本题有很多变形，例如三个瓶子换一瓶汽水，四个瓶子换一瓶汽水等等
#define Exchange 2

int DrinkingProblem(int bottle) {
    int ret = bottle;
    while (bottle >= Exchange) {
        ret += bottle / Exchange;
        bottle = bottle / Exchange + bottle % Exchange;
    }
    return ret;
}

int main() {
    //让用户输入用多少钱来买汽水
    printf("输入钱数：");
    int money = 0;
    scanf("%d", &money);
    int ret = DrinkingProblem(money);
    printf("%d\n", ret);
    return 0;
}
//本体思想上还算简单，需要控制变量的变化
//本代用#define宏定义了交换的比例Exchange，本题是两瓶换一瓶汽水，例如可以换成四瓶换一瓶汽水

 

//第二次尝试：
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
//本题有很多变形，例如三个瓶子换一瓶汽水，四个瓶子换一瓶汽水等等
#define Exchange 2

int DrinkingProblem(int bottle) {
    int ret = bottle;
    while (bottle >= Exchange) {
        ret += bottle / Exchange;
        bottle = bottle / Exchange + bottle % Exchange;
    }
    return ret;
}


//这可以看成是一个等差数列
int DrinkingProblem1(int bottle) {
    if (bottle <= 0) {
        return 0;
    }
    else {
        int ret1 = 0;
        return bottle * 2 - 1;
    }
}

int main() {
    //让用户输入用多少钱来买汽水
    printf("输入钱数：");
    int money = 0;
    scanf("%d", &money);
    int ret = DrinkingProblem(money);
    int ret1 = DrinkingProblem1(money);
    printf("%d %d\n", ret, ret1);
    return 0;
}
//此问题可以看成是等差数列，所以使用公式即可求得