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