Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer x and k,
and tells Arya k but not x.
Arya have to find the value .
There are n ancient numbers c1, c2, ..., cn and
Pari has to tell Arya
if
Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or
not. Formally, is it true that Arya can understand the value
for
any positive integer x?
Note, that means
the remainder of x after dividing it by y.
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.
4 5 2 3 5 12
Yes
2 7 2 3
No
In the first sample, Arya can understand because 5 is
one of the ancient numbers.
In the second sample, Arya can't be sure what is.
For example 1 and 7 have
the same remainders after dividing by 2 and 3,
but they differ in remainders after dividing by 7.
题目大意:
给你n个数ci,和一个k,已知x mod ci,问你能不能确定x mod k
解法:
等于给你n个数,让你求x,然后确定x mod k是不是同一个数
中国剩余定理,唔贴个链接,这个博客说的反正我能看懂。。。
http://blog.csdn.net/acdreamers/article/details/8050018
然后其实最后我们得到的结论就是只要k 能整除 ci的最小公倍数那么就可以得到x mod k确定
以下是代码。
#include <cstdio> using namespace std; typedef long long LL; LL gcd(LL a, LL b){ while(b){ int tmp = a % b; a = b; b = tmp; } return a; } int main() { LL n, k, c, lcm = 1; scanf("%I64d%I64d", &n, &k); for(int i = 0; i < n; ++i){ scanf("%I64d", &c); lcm = (c * lcm) / gcd(lcm, c); lcm %= k; } if(lcm == 0) printf("Yes "); else printf("No "); return 0; }