题解 | 汽水瓶

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func main() {
	scanner := bufio.NewScanner(os.Stdin)
	for {
		if scanner.Scan() {
            resCount := 0
			emptyCountStr := scanner.Text()
			emptyCountInt, _ := strconv.Atoi(emptyCountStr)
			if emptyCountInt == 0 {
				break
			}
			if emptyCountInt < 0 {
				resCount = 0
			} else if emptyCountInt == 2 {
				resCount = 1
			} else {
				for {
					canDrink := emptyCountInt / 3
					resCount += canDrink
					emptyCountRelease := emptyCountInt % 3
					emptyCountInt = canDrink + emptyCountRelease
					if emptyCountInt <= 1 {
						break
					} else if emptyCountInt == 2 {
						resCount++
						break
					}
				}
			}
            fmt.Printf("%d\n", resCount)

		}
	}
	
}

解题思路：当剩余的汽水瓶是1的时候，无法再通过借来兑换，当剩余的为2时，可以通过借空瓶子来集齐3个空瓶，再兑换一瓶喝掉然后归还。因此循环时只需要注意剩余的汽水瓶到达临界点时的处理条件即可