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  • Codeforces Round #360 (Div. 2) D. Remainders Game

    D. Remainders Game
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Today Pari and Arya are playing a game called Remainders.

    Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value . There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya  if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value  for any positive integer x?

    Note, that  means the remainder of x after dividing it by y.

    Input

    The first line of the input contains two integers n and k (1 ≤ n,  k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

    The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).

    Output

    Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

    Examples
    input
    4 5
    2 3 5 12
    output
    Yes
    input
    2 7
    2 3
    output
    No
    Note

    In the first sample, Arya can understand  because 5 is one of the ancient numbers.

    In the second sample, Arya can't be sure what  is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

     题意:给你n个数,一个k;可以告诉你xmod ci的值;求x%k是否唯一;

     思路:根据中国剩余定理,如果中国剩余定理有解x,另外一个解为x+lcm(c0,c1...cn);

        所以lcm%k==0;

    #include<bits/stdc++.h>
    using namespace std;
    #define ll __int64
    #define mod 1000000007
    #define pi (4*atan(1.0))
    const int N=1e3+10,M=1e6+10,inf=1e9+10;
    ll gcd(ll a,ll b)
    {
        return b==0?a:gcd(b,a%b);
    }
    int main()
    {
        ll x,y,z,i,t;
        ll lcm=1;
        scanf("%lld%lld",&x,&y);
        for(i=0;i<x;i++)
        {
            scanf("%lld",&z);
            lcm=z*lcm/gcd(z,lcm);
            lcm%=y;
        }
        if(lcm==0)
        printf("Yes
    ");
        else
        printf("No
    ");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5629186.html
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