codeforces 338D GCD Table （扩展中国剩余定理）

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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 55 2 1 2 1
output
YES
input
100 8 55 2 1 2 1
output
NO
input
100 100 71 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[N],c[N],r[N];  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");}

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