题目名称:
喝汽水问题
题目内容:
喝汽水,1瓶汽水1元,2个空瓶可以换一瓶汽水,给20元,可以多少汽水(编程)
分析:
1.20元能喝20瓶
2.2个空瓶能换1瓶汽水,剩下的空瓶:empty/2+empty%
代码
方法1:
#include<stdio.h>
int main()
{
int money = 0;
int total = 0;
int empty = 0;
scanf("%d", &money);
total = money;
empty = money;
while (empty > 1)
{
total += empty / 2;
empty = empty / 2 + empty % 2;
}
printf("%d", total);
return 0;
}
方法2:(由方法1得等差数列(empty*2-1))
#include<stdio.h>
int main()
{
int money = 0;
int total = 0;
int empty = 0;
scanf("%d", &money);
total = money;
empty = money;
if (empty > 0)
total = empty * 2 - 1;
return 0;
}
感想:
没有撤退可言
标签:汽水,--,money,C语言,int,空瓶,total,empty From: https://blog.51cto.com/u_15501985/5916391