(相关资料图)
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:
#includeint 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))
#includeint 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