Problem Description
source:https://uva.onlinejudge.org/external/118/11877.html
Once upon a time, there is a special coco-cola store. If you return three empty bottles to the shop, you’ll get a full bottle of coco-cola to drink. If you have n empty bottles right in your hand, how many full bottles of coco-cola can you drink?
Input
There will be at most 10 test cases, each containing a single line with an integer n (1 ≤ n ≤ 100). The input terminates with n = 0, which should not be processed.
Output
For each test case, print the number of full bottles of coco-cola that you can drink. Spoiler Let me tell you how to drink 5 full bottles with 10 empty bottles: get 3 full bottles with 9 empty bottles, drink them to get 3 empty bottles, and again get a full bottle from them. Now you have 2 empty bottles. Borrow another empty bottle from the shop, then get another full bottle. Drink it, and finally return this empty bottle to the shop!
Sample Input
3
10
81
0
Sample Output
1
5
40
source:https://uva.onlinejudge.org/external/118/11877.html
Once upon a time, there is a special coco-cola store. If you return three empty bottles to the shop, you’ll get a full bottle of coco-cola to drink. If you have n empty bottles right in your hand, how many full bottles of coco-cola can you drink?
Input
There will be at most 10 test cases, each containing a single line with an integer n (1 ≤ n ≤ 100). The input terminates with n = 0, which should not be processed.
Output
For each test case, print the number of full bottles of coco-cola that you can drink. Spoiler Let me tell you how to drink 5 full bottles with 10 empty bottles: get 3 full bottles with 9 empty bottles, drink them to get 3 empty bottles, and again get a full bottle from them. Now you have 2 empty bottles. Borrow another empty bottle from the shop, then get another full bottle. Drink it, and finally return this empty bottle to the shop!
Sample Input
3
10
81
0
Sample Output
1
5
40
Solution:
#include<stdio.h> int main() { int n, get; while(scanf("%d",&n)==1) { if(n==0) break; get=0; while(n>=3) { /* get+=n/3; n=n/3+n%3;*/ n=(n-3)+1; get+=1; } if(n==2) get+=1; printf("%d\n",get); } return 0; }
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