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.
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).
Print a single word "YES", if the given sequence occurs in table G, otherwise print "NO".
100 100 5
5 2 1 2 1
YES
100 8 5
5 2 1 2 1
NO
100 100 7
1 2 3 4 5 6 7
NO
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.
数学问题 扩展中国剩余定理
首先可以知道这个数列如果存在，它最早会出现在“所有a[]的lcm”这一行，而如果lcm行都没有，之后的行每个数都成倍扩大，更不可能出现目标数列。
我们假设数列第一个元素在第x列，因为是在gcd表上，所以a[i]一定是x的因数，也就是$ x = a[i]*k $
这个式子的形式看上去很熟悉？ $ x equiv 0 (mod a[i])$
像这样继续把每个式子都列出来：
$x equiv 0 (mod a_{1})$
$x+1 equiv 0 (mod a_{2})$
$x+2 equiv 0 (mod a_{3})$
$x+3 equiv 0 (mod a_{4})$
$x+4 equiv 0 (mod a_{5})$
这样就得到了一串$x equiv a[i]-i+1 (mod a_{i})$ 形式的同余方程。
然后用中国剩余定理解就行……吗？注意到这些b[]可能并不是互质的，我们需要扩展中国剩余定理
扩展中国剩余定理？听起来好帅
实际上就是用扩展欧几里得合并中国剩余定理，消除不互质的模数之间的gcd的影响。
我们知道中国剩余定理就是用扩展欧几里得算法推出来的，那么它实际上就是扩展·扩展·扩展欧几里得算法？
一连串花式化简好麻烦的，用latex写公式更麻烦。
推荐左转http://blog.csdn.net/clove_unique/article/details/54571216
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 #define LL long long 7 using namespace std; 8 const int mxn=10010; 9 LL read(){ 10 LL x=0,f=1;char ch=getchar(); 11 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 13 return x*f; 14 } 15 LL ksmul(LL a,LL b,LL p){ 16 LL res=0; 17 if(b<0)b=-b,a=-a; 18 while(b){ 19 if(b&1)(res+=a)%=p; 20 a<<=1;a%=p; 21 b>>=1; 22 } 23 return res; 24 } 25 LL gcd(LL a,LL b){ 26 return (!b)?a:gcd(b,a%b); 27 } 28 LL exgcd(LL a,LL b,LL &x,LL &y){ 29 30 if(!b){x=1;y=0;return a;} 31 LL tmp=exgcd(b,a%b,x,y); 32 LL c=x;x=y;y=c-a/b*y; 33 return tmp; 34 } 35 LL inv(LL a,LL b){ 36 LL x,y; 37 exgcd(a,b,x,y); 38 x=(x%b+b)%b; 39 if(!x)x+=b; 40 return x; 41 } 42 LL n,M,k,lcm; 43 LL a[mxn],m[mxn],c[mxn]; 44 int main(){ 45 int i,j; 46 n=read();M=read();k=read(); 47 for(i=1;i<=k;i++)a[i]=read(); 48 lcm=a[1]; 49 for(i=2;i<=k;i++){ 50 lcm=lcm/gcd(lcm,a[i])*a[i]; 51 if(lcm>n){printf("NO ");return 0;} 52 } 53 for(i=1;i<=k;i++)m[i]=a[i],c[i]=a[i]-i+1LL; 54 LL m1=m[1],c1=c[1]; 55 for(i=2;i<=k;i++){ 56 LL m2=m[i],c2=c[i]; 57 LL tmp=gcd(m1,m2); 58 if((c2-c1)%tmp){printf("NO ");return 0;}//无解 59 LL m3=m1/tmp*m2,c3=inv(m1/tmp,m2/tmp); 60 c3=ksmul(c3,(c2-c1)/tmp,m2/tmp); 61 c3=ksmul(c3,m1,m3); 62 c3=((c3+c1)%m3+m3)%m3; 63 m1=m3;c1=c3; 64 } 65 if(!c1)c1+=m1; 66 if(c1>M-k+1){printf("NO ");return 0;} 67 for(int i=1;i<=k;i++){ 68 LL tmp=gcd(lcm,c1+i-1); 69 if(tmp!=a[i]){printf("NO ");return 0;} 70 } 71 printf("YES "); 72 return 0; 73 }