cdoj 65 CD Making 水题

CD Making

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://acm.uestc.edu.cn/#/problem/show/65

Description

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

2
5 2
13 13

Sample Output

3

2

HINT

题意

题解：

讨论讨论，麻烦……

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200000
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=;if(!x){putchar('');puts("");return;}
    while(x>)CH[++Num]=x%,x/=;
    while(Num)putchar(CH[Num--]+);
    puts("");
}
//**************************************************************************************

int main()
{
    //test;
    int T; cin >> T;
    while (T--) {
        int N, K; scanf("%d%d",&N,&K);
        if (K == ) K = ;
        if (K < ) printf("%d\n",N/K + (N % K ?  : ));
        else if (N == ) printf("2\n");
        else if (K >= ) printf("%d\n",N/K + (N % K ?  : ));
        else if (K == ) {
            int r = N % K;
            if (r != ) printf("%d\n",N / K + (r ==  ?  : ));
            else printf("%d\n",N / K + );
        }
    }
}