NOIP 模拟 $11; m math$

题解

签到题（然而还是不会）

考虑所有可能的值一定是 (in [0,k))，且一定为 (gcd(a_1,a_2,...a_n,k)) 的倍数。

证明：

设 (tmp=b_1a_1+b_2a_2+...b_na_n) 那么 (tmp) 可表示为 (k_1×gcd(tmp,k))， (k) 可表示为 (k_2×gcd(tmp,k))

故 (tmp;;mod;; k) 也为 (gcd(tmp,k)) 的倍数

证毕

Code:

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
}
using IO::read;
namespace nanfeng{
    #define cmax(x,y) ((x)>(y)?(x):(y))
    #define cmin(x,y) ((x)>(y)?(y):(x))
    #define FI FILE *IN
    #define FO FILE *OUT
    static const int N=5e5+7;
    int num[N],n,k;
    int gcd(int a,int b) {return b?gcd(b,a%b):a;}
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        read(n),read(k);
        for (ri i(1);i<=n;p(i)) read(num[i]);
        int g=k;
        for (ri i(1);i<=n;p(i)) g=gcd(g,num[i]);
        printf("%d
",k/g);
        for (ri i(0);i<k;i+=g) printf("%d ",i);
        return 0;
    } 
}
int main() {return nanfeng::main();}