CodeCraft20 (Div. 2) C. Primitive Primes

题意：

f(x) = a0+a1*x+  +an*x^(n-1) , g(x) = b0+b1*x+  +bn*x^(n-1)，h(x) = f(x)*g(x)，问h（x）的哪一项系数模p!=0

题解：

h[i+j]=sigma{a[i]*b[j]}

a[]%p={0,0,0,非0,..}

b[]%p={0,0,非0,..}

ans=两个非0第一次出现的下标之和。

（a[i]*b[j]）%p=a[i]%p*b[j]%p!=0

因为h[i+j]=a[i]*b[j]+a[i-1]*b[j+1]+..a[i+1]*b[j-1]+...

                  非0        0                               0

代码

#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define check system("pause")
#define all(x) (x).begin(),(x).end()
#define de(a) cout<<#a<<" = "<<a<<endl
#define dd(a) cout<<#a<<" = "<<a<<" "
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define lowbit(a) ((a)&-(a))
#define INF 0x3f3f3f3f
const ll mod = 1e9+7;
const int N = 1e6+20;
#define dep(i,a,b) for(int i=(a);i>=(b);i--)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define mes(p,b) memset(p,b,sizeof(p))
#define sz(x) int(x.size())
int n,m,x,p,ind1=-1,ind2=-1; 
int main()
{
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
     cin>>n>>m>>p;
    rep(i,0,n-1){
        cin>>x;
        if(ind1==-1&&x%p!=0) ind1=i;
    } 
    rep(i,0,m-1){
        cin>>x;
        if(ind2==-1&&x%p!=0) ind2=i;
    }
    cout<<ind1+ind2; 
      return 0;
}