CodeForces 858B Which floor?

最新推荐文章于 2019-11-07 13:12:04 发布

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最新推荐文章于 2019-11-07 13:12:04 发布

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本文介绍了一个算法问题，通过已知部分楼层与房间号的关系，来推断特定房间位于哪一层。该算法考虑了不同情况下的特殊情况，并通过输入输出示例帮助理解。

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题目:

In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1.

Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers.

Given this information, is it possible to restore the exact floor for flat n?

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory.

m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≤ ki ≤ 100, 1 ≤ fi ≤ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct.

It is guaranteed that the given information is not self-contradictory.

Output

Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor.

   Examples 
 

     input 
   
10 3
6 2
2 1
7 3

     output 
   
4

     input 
   
8 4
3 1
6 2
5 2
2 1

     output 
   
-1

Note

In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor.

In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat.

代码:

#include<stdio.h>
#include<algorithm>
using namespace std;

int main()
{
    int m,n,i,x,y,mi=0,ma=105,k=0,ans,flag=1;
    scanf("%d%d",&m,&n);
    if(n==0)
    {
        if(m==1)printf("1\n");                  //特判m=1,n=0
        else printf("-1\n");                    //特判m>1,n=0
    }
    else
    {
        for(i=0;i<n;i++)                        //x为已知房间数 y为楼层
        {                                       //y-1楼最大值为x-1 y楼最小值为x
            int w;
            scanf("%d%d",&x,&y);                //接下来找到最大值里的最小值,最小值里的最大值
            if(y==1)                            //防止除0
                if(x>mi)mi=x;
            else
            {
                w=(x-1)/(y-1);                  //上一层最大值
                if(w<ma)ma=w;                   //找到最大值里的最小值存到ma

                w=0;                            //除不尽结果要+1
                if(x%y!=0)w++;
                w+=x/y;                         //找到最小值里的最大值存到mi
                if(w>mi)mi=w;
            }
        }
        if(m%mi!=0)k++;                         //确定可能楼层是否唯一
        k+=m/mi;
        ans=k;
        k=0;
        if(m%ma!=0)k++;
        k+=m/ma;
        if(ans!=k)flag=0;
        if(flag==1)
            printf("%d\n",ans);
        else
            printf("-1\n");
    }
    return 0;
}