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 banknote denominations on Mars: the value of -th banknote is . Natasha has an infinite number of banknotes of each denomination.
Martians have fingers on their hands, so they use a number system with base . In addition, the Martians consider the digit (in the number system with base ) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base is , the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
The first line contains two integers and (
The second line contains integers () — denominations of banknotes on Mars.
All numbers are given in decimal notation.
On the first line output the number of values 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.
2 8
12 20
2
0 4
3 10
10 20 30
1
0
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of , you will get in octal system. The last digit is .
If you take one banknote with the value of and one banknote with the value of , the total value will be . In the octal system, it is . The last digit is .
If you take two banknotes with the value of , the total value will be , this is in the octal system. The last digit is .
No other digits other than and can be obtained. Digits and could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
题解:
裴蜀定理板题,然而当时玩交互题去了,结果最后没时间打了......
先求出所有数模k后的gcd,由裴蜀定理,(gcd*i)%k(i=1,2,3,......k)必然是其中一个答案,枚举i即可
注意对0的处理
#include<bits/stdc++.h> using namespace std; int n,k,ans,gcd=-1,num; bool mk=false; bool vis[100005]; int get_gcd(int a,int b) { return b==0?a:get_gcd(b,a%b); } int main() { memset(vis,false,sizeof(vis)); scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) { scanf("%d",&num); num=num%k; if(!num) mk=true; if(num&&gcd!=-1) gcd=get_gcd(gcd,num); if(num&&gcd==-1) gcd=num; } if(gcd==-1) printf("1 0"); else { int tot=0; if(mk) { tot++; vis[0]=true; } for(int i=1;i<=k;i++) if(!vis[(i*(long long)gcd)%(long long)k]) { tot++; vis[(i*(long long)gcd)%(long long)k]=true; } printf("%d ",tot); for(int i=0;i<k;i++) if(vis[i]) printf("%d ",i); } return 0; }