codeforces 1316 C math

不会

题意：给出了两个多项式的系数，求两个多项式相乘后问系数不能被 p 整除的幂的值。

思路：ci​=a∗bi​+a1​∗bi−1​+…+ai−1​∗b1​+ai​∗b0​。

　　倘若

 如果在把两个多项式乘开之后，满足c%p！=0，则一定有一个和c相同x幂的系数（a*b)%p != 0,但是如果还有其他相同x幂的系数爷满足不整除0，则结果c不一定为整除p。

如果取a，b分别为a数组和b最先不能整除p的数，因为和(a*b)具有相同x幂的系数是有（a前面的数*b后面的数)或者（b前面的数*a后面的数)，因为a，b前面都没有不能整除p的数。所以除了a*b满足不能整除p，其余的都能整除p，所以c%p != 0。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 1e6+10;
ll n,m,p;
inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}
ll a[maxn];
ll b[maxn];
int main()
{
    n = read();
    m = read();
    p = read();
    rep(i,0,n-1) {
        a[i] = read();
        a[i] %= p;
    }
    rep(i,0,m-1) {
        b[i] = read();
        b[i] %= p;
    }
    ll pos1 =0,pos2 = 0;
    rep(i,0,n-1){
        if(a[i] != 0) {
            pos1 = i;
            break;
        }
    }
    rep(i,0,m-1){
        if(b[i] != 0) {
            pos2 = i;
            break;
        }
    }
    printf("%lld\n",pos1+pos2 );

}

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来源： https://www.cnblogs.com/jrfr/p/13223094.html