Codeforce-CodeCraft-20 (Div. 2)-C. Primitive Primes（本原多项式+数学推导）
It is Professor R’s last class of his teaching career. Every time Professor R taught a class, he gave a special problem for the students to solve. You being his favourite student, put your heart into solving it one last time.
You are given two polynomials f(x)=a0+a1x+⋯+an−1xn−1 and g(x)=b0+b1x+⋯+bm−1xm−1, with positive integral coefficients. It is guaranteed that the cumulative GCD of the coefficients is equal to 1 for both the given polynomials. In other words, gcd(a0,a1,…,an−1)=gcd(b0,b1,…,bm−1)=1. Let h(x)=f(x)⋅g(x). Suppose that h(x)=c0+c1x+⋯+cn+m−2xn+m−2.
You are also given a prime number p. Professor R challenges you to find any t such that ct isn’t divisible by p. He guarantees you that under these conditions such t always exists. If there are several such t, output any of them.
As the input is quite large, please use fast input reading methods.
Input
The first line of the input contains three integers, n, m and p (1≤n,m≤106,2≤p≤109), — n and m are the number of terms in f(x) and g(x) respectively (one more than the degrees of the respective polynomials) and p is the given prime number.
It is guaranteed that p is prime.
The second line contains n integers a0,a1,…,an−1 (1≤ai≤109) — ai is the coefficient of xi in f(x).
The third line contains m integers b0,b1,…,bm−1 (1≤bi≤109) — bi is the coefficient of xi in g(x).
Output
Print a single integer t (0≤t≤n+m−2) — the appropriate power of x in h(x) whose coefficient isn’t divisible by the given prime p. If there are multiple powers of x that satisfy the condition, print any.
Examples
inputCopy
3 2 2
1 1 2
2 1
outputCopy
1
inputCopy
2 2 999999937
2 1
3 1
outputCopy
2
Note
In the first test case, f(x) is 2x2+x+1 and g(x) is x+2, their product h(x) being 2x3+5x2+3x+2, so the answer can be 1 or 2 as both 3 and 5 aren’t divisible by 2.
In the second test case, f(x) is x+2 and g(x) is x+3, their product h(x) being x2+5x+6, so the answer can be any of the powers as no coefficient is divisible by the given prime.
题意：
给定两个多项式长度 n 和 m ，再给定每一项的系数，由常数项到最高次项排序，其中每个多项式的系数的GCD=1。
然后再给定一个质数 p
问两个多项式相乘后得到的第三个多项式中，哪一项的系数不是 p 的倍数，输出这个项的x的幂次（下标）
如果，那么有且，因为幂是低次幂向高次幂排列，乘积也是如此，因此我们只要找到最小非0次幂不能整除P，即可。即找到最小的m和n即可。
是第n+m项
可行性：本原多项式
如果还有问题的话，欢迎DL补充，小弟不胜感激，洗耳恭听。
#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
char ch = getchar();
x = 0;
t f = 1;
while (ch < '0' || ch > '9')
f = (ch == '-' ? -1 : f), ch = getchar();
while (ch >= '0' && ch <= '9')
x = x * 10 + ch - '0', ch = getchar();
x *= f;
}
#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 200005
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
int main()
{
int m, n, p, tem;
read(n), read(m), read(p);
ll ans1 = 0, ans2 = 0;
for (int i = 0; i < n; i++)
{
read(tem);
tem %= p;
if (tem && !ans1)
ans1 = i;
}
for (int i = 0; i < m; i++)
{
read(tem);
tem %= p;
if (tem && !ans2)
ans2 = i;
}
cout << ans1 +ans2 << endl;
}
写在最后:
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