TOJ 3976.Change（dp或找规律）

题目链接：http://acm.tju.edu.cn/toj/showp3976.html

3976.   Change

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 530   Accepted Runs: 244

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After Shawn sees the following picture, he decides to give up his career in IT and turn to sell fruits. Since Shawn is a lazy guy, he doesn’t like to do any extra work. When he starts selling fruits, he finds that he always needs to take changes for customers. He wants you to write a program to give the least number of changes. Shawn is also a strange guy, because he takes changes by 6,5,3,1.

Input

The first line of input is an integer T which indicates the sum of test case. Each of the following T lines contains an integer n(1≤n≤10⁵), which is the total money Shawn need to give to the customer.

Output

For each test case, output "Case i: Result" in one line where i is the case number and Result is the least number of changes Shawn need to give customer.

Sample Input

341019

Sample output

Case 1: 2Case 2: 2Case 3: 4

Hint

Case 1: 4=3+1Case 2: 10=5+5Case 3: 19=6+6+6+1

Source: TJU Team Selection 2013

 

找零钱问题，本以为要用dp，但貌似找规律也是一个办法，不过规律不好找，1~6比较特殊单独考虑，随后从7开始进入规律的循环了，自己推一下：从7开始每6个是一个循环，也就是加了个6，知道这些写起来就非常简单了。

 

#include <stdio.h>int a[12]={1,2,1,2,1,1,2,2,2,2,2,2};int main(){int t,n,sum;scanf("%d",&t);for(int i=0;i<t;i++){scanf("%d",&n);sum=0;while(n>12){n-=6;sum++;}sum+=a[n-1];printf("Case %d: %d\n",i+1,sum); }}