hdu3892 Common Roots 最大取模公因式

We have many polynomials modulo p (p is a prime number). An interesting issue would be to determine whether they have some roots in common. Notice roots we mention here are integers in modulo p system (0 <= root < p). Moreover, if
the given polynomial is of order r, we will guarantee that it has r roots.

For example, we have

x^2 + 13x + 36 (mod 37)

x^3 + 14x^2 + 49x + 36 (mod 37)

If x = 33 or x = 28, both of them would give the value of 0. So 33 and 28 are the roots in common.

[align=left]Input[/align]
There are many test cases (less than1000).

In each case, the integer in the first line is n (the number of polynomials in this case). Then n lines followed. Each of them starts with an integer r (order of polynomials, r <= 50), and r + 1 integers (a(r), a(r-1) ,..., a(0)), which means the polynomial
goes like:

a(r) * x^r + a(r-1) * x^(r-1) + … +a(1) * x + a(0) (mod 999983).

To make it easier, p is set to be 999983, as you see.

[align=left]Output[/align]
For each case, just output “YES” if they have common roots, otherwise “NO” in a single line.

[align=left]Sample Input[/align]

2
2 1 13 36
3 1 14 49 36

[align=left]Sample Output[/align]

YES

是否存在解，满足每个式子取模后都为0。

求出这些式子的公因式，如果最大系数大于0的话肯定有解。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
#define eps 1e-9
#define MAXN 1010
#define MAXM 110
#define MOD 999983
typedef long long LL;
using namespace std;
LL N,M;
vector<LL> ans,g[MAXN];
void gcd(LL a,LL b,LL& d,LL& x,LL& y){
if(!b){
d=a;
x=1;
y=0;
}
else{
gcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
LL inv(LL a,LL n){
LL d,x,y;
gcd(a,n,d,x,y);
return d==1?(x+n)%n:-1;
}
vector<LL> poly_gcd(vector<LL> a,vector<LL> b){
if(!b.size()) return a;
int L=a.size()-b.size()+1,bs=b.size(),as=a.size();
for(LL i=0;i<L;i++){
LL t=(a[i]*inv(b[0]%MOD,MOD))%MOD;
for(LL j=0;j<bs;j++) a[i+j]=(a[i+j]-b[j]*t%MOD+MOD)%MOD;
}
vector<LL> c;
LL flag=0;
for(LL i=0;i<as;i++){
if(a[i]){
flag=1;
for(LL j=i;j<as;j++) c.push_back(a[j]);
}
if(flag) break;
}
return poly_gcd(b,c);
}
int main(){
freopen("in.txt","r",stdin);
LL cas=0;
while(scanf("%I64d",&N)!=EOF){
for(LL i=0;i<N;i++) g[i].clear();
for(LL i=0;i<N;i++){
scanf("%I64d",&M);
LL t;
for(LL j=0;j<M+1;j++){
scanf("%I64d",&t);
g[i].push_back(t);
}
}
if(N==1) ans=g[0];
else{
ans=poly_gcd(g[0],g[1]);
for(LL i=2;i<N;i++) ans=poly_gcd(ans,g[i]);
}
if(ans.size()>1) puts("YES");
else puts("NO");
}
return 0;
}