codeforces 338D GCD Table （扩展中国剩余定理）

D. GCD Table

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Consider a table G of size n × m such
that G(i, j) = GCD(i, j) for all 1 ≤ i ≤ n, 1 ≤ j ≤ m. GCD(a, b) is
the greatest common divisor of numbers a and b.

You have a sequence of positive integer numbers a1, a2, ..., ak.
We say that this sequence occurs in table G if it coincides with consecutive elements in some row, starting from some position. More
formally, such numbers 1 ≤ i ≤ n and 1 ≤ j ≤ m - k + 1 should
exist that G(i, j + l - 1) = al for
all 1 ≤ l ≤ k.

Determine if the sequence a occurs in table G.

Input

The first line contains three space-separated integers n, m and k (1 ≤ n, m ≤ 1012; 1 ≤ k ≤ 10000).
The second line contains k space-separated integers a1, a2, ..., ak (1 ≤ ai ≤ 1012).

Output

Print a single word "YES", if the given sequence occurs in table G,
otherwise print "NO".

Examples

input

100 100 5
5 2 1 2 1

output

YES

input

100 8 5
5 2 1 2 1

output

NO

input

100 100 7
1 2 3 4 5 6 7

output

NO

Note

Sample 1. The tenth row of table G starts from sequence {1, 2, 1, 2, 5, 2, 1, 2, 1, 10}. As you can see, elements from fifth to ninth
coincide with sequence a.

Sample 2. This time the width of table G equals 8. Sequence a doesn't
occur there.

题目大意：给出一个n*m的数表，(i,j)=GCD(i,j).然后给出一个序列a1...ak问序列是否在数表中出现过。

题解：扩展中国剩余定理

首先可以确定行一定是a1...ak的最小公倍数的倍数，如果lcm>n，那么无解。

然后设第一列为x 

x=a1*b1  (其中b表示a1的整数倍，因为a1为x的gcd，所以一定能表示成a1*b1的形式)

x+1=a2*b2

x+2=a3*b3

.....

可以把式子都转换成线性同余方程的形式x=a[i]-i+1 (mod a[i])

用扩展中国剩余定理合并，然后求出解x

如果解出来的x为0，那么需要先加上r，再进行判断。

如果x>m-k+1,则无解。

然后再带入验证一下答案，就可以输出最终判断的结果了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define N 10003
#define LL long long
using namespace std;
LL n,m,a
,c
,r
;
int k;
LL mul(LL a,LL b,LL mod)
{
LL ans=0;
while (b) {
if (b&1) ans=(ans+a)%mod;
b>>=1;
a=(a+a)%mod;
}
return ans%mod;
}
LL gcd(LL x,LL y)
{
LL r;
while (y) {
r=x%y;
x=y; y=r;
}
return x;
}
void exgcd(LL a,LL b,LL &x,LL &y)
{
if (!b) {
x=1; y=0; return;
}
exgcd(b,a%b,x,y);
LL t=y;
y=x-(a/b)*y;
x=t;
}
LL inv(LL a,LL b){
LL x,y;
exgcd(a,b,x,y);
return (x%b+b)%b;
}
bool check(LL a1,LL a2,LL r1,LL r2,LL &aa,LL &rr)
{
LL c=a2-a1; LL d=gcd(r1,r2);
// cout<<r1<<" "<<r2<<" "<<d<<endl;
if (c%d) return 0;
c/=d; r1/=d; r2/=d;
LL x=inv(r1,r2); c=(c%r2+r2)%r2;
rr=r1*r2*d;
x=mul(x,c,r2);
x=mul(mul(x,r1,rr),d,rr)+a1;
aa=(x%rr+rr)%rr;
return 1;
}
int main()
{
freopen("a.in","r",stdin);
scanf("%I64d%I64d%d",&n,&m,&k);
for (int i=1;i<=k;i++) scanf("%I64d",&c[i]);
LL lcm=c[1]/gcd(c[1],c[2])*c[2];
for (int i=3;i<=k;i++){
lcm=lcm/gcd(lcm,c[i])*c[i];
if (lcm>n) {
printf("NO\n");
return 0;
}
}
for (int i=1;i<=k;i++) r[i]=c[i],a[i]=c[i]-i+1;
LL a1,a2,r1,r2,rr,aa;
a1=aa=a[1]; r1=rr=r[1];
for (int i=2;i<=k;i++) {
a2=a[i]; r2=r[i];
if (a2<0) a2=(a2%r2+r2)%r2;
if(!check(a1,a2,r1,r2,aa,rr)) {
printf("NO\n");
return 0;
}
a1=aa; r1=rr;
//cout<<aa<<" "<<rr<<endl;
}
//cout<<aa<<endl;
if (!aa) aa+=rr;
if (aa>m-k+1) {
printf("NO\n");
return 0;
}
for (int i=1;i<=k;i++) {
LL t=gcd(lcm,aa+i-1);
if (t!=c[i]) {
printf("NO\n");
return 0;
}
}
printf("YES\n");
}