CodeForces 687B Remainders Game

数论。

如果$x$不唯一，假设存在两个解，较大的为${x_1}$，较小的为${x_2}$。

那么，

$\left\{ {\begin{array}{*{20}{c}}
{{x_1}\% {c_i} = {x_2}\% {c_i}}\\
{{x_1}\% k \ne {x_2}\% k}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{({x_1} - {x_2})\% {c_i} = 0}\\
{({x_1} - {x_2})\% k \ne 0}
\end{array}} \right.$。

$∵lcm({c_1},{c_2},{c_3},......,{c_n})\% {c_i} = 0$

$∴lcm({c_1},{c_2},{c_3},......,{c_n})\% ({x_1} - {x_2}) = 0$

$∵({x_1} - {x_2})\% k \ne 0$

$∴lcm({c_1},{c_2},{c_3},......,{c_n})\% k \ne 0$

也就是说：如果解不唯一，那么$lcm({c_1},{c_2},{c_3},......,{c_n})\% k \ne 0$。

换句话说就是：如果解唯一，那么$lcm({c_1},{c_2},{c_3},......,{c_n})\% k = 0$。

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<iostream>

using namespace std;

typedef long long LL;

const double pi=acos(-1.0),eps=1e-;

void File()

{

    freopen("D:\\in.txt","r",stdin);

    freopen("D:\\out.txt","w",stdout);

}

template <class T>

inline void read(T &x)

{

    char c = getchar(); x = ;while(!isdigit(c)) c = getchar();

    while(isdigit(c)) { x = x *  + c - ''; c = getchar();  }

}

LL gcd(LL a,LL b)

{

    if(b==) return a;

    return gcd(b,a%b);

}

LL lcm(LL a,LL b)

{

    return a*b/gcd(a,b);

}

int n;

LL ans=,k;

int main()

{

    scanf("%d%lld",&n,&k);

    for(int i=;i<=n;i++)

    {

        LL x; scanf("%lld",&x);

        ans=lcm(ans,x)%k;

    }

    if(ans) printf("No\n");

    else printf("Yes\n");

    return ;

}

巴特西

CodeForces 687B Remainders Game

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