[Codeforces 1011E] Border

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[題目連結]

        https://codeforces.com/contest/1011/problem/E

[演算法]

         裴蜀定理 ： 設為n個整數，d是它們的最大公約數，那麼存在整數 使得 

         顯然 ， 我們只需求出a1,a2...an模k意義下的最大公約數G，然後枚舉G的倍數即可

         時間複雜度 ： O(NlogK)

[代碼]

        

#include<bits/stdc++.h>using namespace std;#define MAXN 200010int n , k;int a[MAXN];template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }template <typename T> inline void read(T &x){    T f = 1; x = 0;    char c = getchar();    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;    x *= f;}inline int gcd(int x,int y){        if (y == 0) return x;        else return gcd(y,x % y);}int main(){                read(n); read(k);         for (int i = 1; i <= n; i++)         {                read(a[i]);                a[i] %= k;                if (a[i] == 0) a[i] = k;        }        int g = a[1];        for (int i = 2; i <= n; i++) g = gcd(g,a[i]);        set< int > ans;        int now = 0;        for (int i = 0; i < k; i++)         {                ans.insert(now);                now = (now + g) % k;        }        printf("%d\n",(int)ans.size());        for (set< int > :: iterator it = ans.begin(); it != ans.end(); it++) printf("%d ",*it);        printf("\n");                return 0;    }

 

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