UESTC 65 CD Making 贪心法

CD Making

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit 
Status

Tom has N songs
and he would like to record them into CDs. A single CD can contain at most K songs.
In addition, Tom is very superstitious 
and he believes the number 

13

 would
bring bad luck, so he will never let a CD contain exactly 13 songs.
Tom wants to use as few CDs as 
possible to record all these songs. Please help him.

Input

There are T test
cases. The first line gives T,
number of test cases. T lines
follow, each contains N and K,
number of songs Tom wants to 
record into CDs, and the maximum number of songs a single CD can contain.
1≤N≤1000,1≤K≤1000

Output

For each test case, output the minimum number of CDs required, if the above constraints are satisfied.

Sample input and output

Sample Input	Sample Output
2 5 2 13 13	3 2

Hint

If Tom has 5 songs
to record and a single CD can contain up to 2 songs,
Tom has to use 3 CDs
at minimum, each contains 2, 2, 1 songs, 
respectively.
In the second case, Tom wants to record 13 songs,
and a single CD can hold 13 songs
at most. He would have been able to use only 1 CD 
if 
he were not so superstitious. However, since he will not record exactly 13 songs
into a single CD, he has to use 2 CDs
at least, the first 
contains 

12 songs
and the second contains one(Other solutions to achieve 2 CDs
are possible, such as (11, 2),
(10, 3),
etc.).

Source

The 5th UESTC Programming Contest Preliminary

My Solution

秒杀题，就是那个K==14的时候要想到，要额外处理

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
int T,N,K,ans;
scanf("%d",&T);
while(T--){
ans=0;
scanf("%d%d",&N,&K);
if(N<=K) {if(N!=13)printf("1"); else printf("2");}
else {
if(K==13) {ans=N/12;if(N%12!=0) ans+=1;}
else if(K<13||K>14) {
ans=N/K; if(N%K!=0) ans+=1;         //如果K>14&&N%K==13，只要从另外一个地方补个过来变成14就可以了，不用另外加CD
}
else {                                  //K==14
ans=N/K;
if(N%K!=0){
if(N%K!=13) ans+=1;
else ans+=2;
}
}
printf("%d",ans);
}
if(T) printf("\n");
}
return 0;
}

谢谢