E. Border
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are nn banknote denominations on Mars: the value of ii-th banknote is aiai. Natasha has an infinite number of banknotes of each denomination.
Martians have kk fingers on their hands, so they use a number system with base kk. In addition, the Martians consider the digit dd (in the number system with base kk) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base kk is dd, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values dd Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers nn and kk (1≤n≤1000001≤n≤100000, 2≤k≤1000002≤k≤100000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values dd for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
题意：给你n个数，每个数能无限次和其他任意个数自由组合得到一个新的数，所有新的数在k进制下的最后一位有多少种情况，按升序输出所有情况。
思路：预处理出每一个数在k进制下的最后一位的情况，然后得出他们的最大公因数，根据裴蜀定理 得到的最大公因数可以算出他们的在k内的所有可能组合，注意超过的部分要取模。
#include<bits/stdc++.h>
using namespace std;
const int maxn = int(1e5) + 100;
int num[maxn];
int tmp[maxn];
bool vis[maxn];
int gcd(int a,int b) {
return b==0?a:gcd(b,a%b);
}
void solve(int n,int k) {
for(int i=1; i<=n; i++)num[i]%=k;
int comgcd=gcd(num[1],num[2]);
for(int i=3; i<=n; i++)
comgcd=gcd(num[i],comgcd);
sort(num+1,num+n+1);
memset(vis,0,sizeof(vis));
for(int i=1; i<=n; i++) {
for(int j=num[i]; vis[j]==0; j=(j+comgcd)%k)
vis[j]=1;
}
int top=0;
for(int i=0; i<k; i++)
if(vis[i]) tmp[top++]=i;
printf("%d\n",top);
for(int i=0; i<top; i++)
printf("%d%c",tmp[i],i==top-1?'\n':' ');
}
int main() {
int n,k;
while(~scanf("%d %d",&n,&k)) {
for(int i=1; i<=n; i++)
scanf("%d",&num[i]);
solve(n,k);
}
return 0;
}