Oleg and shares (Codeforces

题目链接：

http://codeforces.com/problemset/problem/793/A

A. Oleg and shares

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all n prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices.

Output

Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.

Examples

input

3 312 9 15

output

3

input

2 210 9

output

-1

input

4 11 1000000000 1000000000 1000000000

output

2999999997

题目大意：

输入n、k，然后输入n个数。这n个数中每秒钟有一个数减少k，题目要求计算最少经过几秒钟后，这n个数都相等，如果不存在，那么输出-1 。注意：每秒钟只有一个数减少k。

解题思路：

经过思考会发现，如果存在的话，那么最终结果肯定是 n个数 都等于 最初最小的那个数 。那么先计算出最初最小的那个数，然后计算由剩下的n-1个数变为最小的那个数需要几秒钟即可。如果不存在的话，输出-1.

代码：

#include<iostream>#include<algorithm>using namespace std;int main(){    long long int a[100005],sum,k,z,n;   //sum存储需要的秒数，z只是中间变量    double q;   //q用于进行强制转换，判断能不能被整除    while(cin>>n>>k)    {        sum=0;        z=1;        for(int i=0;i<=n-1;i++)            cin>>a[i];        sort(a,a+n);   //计算最小的那个数，计a[0]        for(int i=1;i<=n-1;i++)        {            q=(double)(a[i]-a[0])/k;   //把整型转化成浮点型，判断能不能整除，如果不能整除，说明不存在            if(q==((a[i]-a[0])/k))   //如果存在，那么把该数字变成最小的那个数 所要经过秒数 加在sum上                sum=sum+(a[i]-a[0])/k;            else   //否则没说明不能被整除，说明该数字无法变为最小的数字，把z标记为0            {                z=0;                break;            }        }        if(z==0)            cout<<-1<<endl;        else            cout<<sum<<endl;    }    return 0;}