HDU 4569 Special equations（枚举+数论）（2013 ACM-ICPC长沙赛区全国邀请赛）

Problem Description

　　Let f(x) = a_nxⁿ +...+ a₁x +a₀, in which a_i (0 <= i <= n) are all known integers. We call f(x) 0 (mod m) congruence equation. If m is a composite, we can factor m into powers of primes and solve every such single equation after which we merge them using the Chinese Reminder Theorem. In this problem, you are asked to solve a much simpler version of such equations, with m to be prime's square.

 

Input

　　The first line is the number of equations T, T<=50.
　　Then comes T lines, each line starts with an integer deg (1<=deg<=4), meaning that f(x)'s degree is deg. Then follows deg integers, representing a_n to a₀ (0 < abs(a_n) <= 100; abs(a_i) <= 10000 when deg >= 3, otherwise abs(a_i) <= 100000000, i<n). The last integer is prime pri (pri<=10000). 
　　Remember, your task is to solve f(x) 0 (mod pri*pri)

 

Output

　　For each equation f(x) 0 (mod pri*pri), first output the case number, then output anyone of x if there are many x fitting the equation, else output "No solution!"

 

题目大意：给你一个最高4次幂的多项式，求一个x，满足f(x) mod phi² = 0。

思路：先枚举x = [0, phi)，如果f(x) mod phi = 0，再枚举x2 = x，每次加phi，直到f(x) mod phi² = 0，输出结果。找不到输出No solution。

PS：我也不知道为什么是对的我看别人说是这么做的……我数论知识很少……我只知道要满足f(x) mod phi² = 0就要先满足f(x) mod phi = 0……

 

代码（62MS）：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <queue>
 using namespace std;
 typedef long long LL;
 
 const int MAXN = ;
 
 int T, deg;
 LL phi;
 LL a[MAXN];
 
 LL f(LL x, LL m) {
     LL ret = , xx = ;
     for(int i = ; i <= deg; ++i) {
         ret = (ret + a[i] * xx) % m;
         xx = (xx * x) % m;
     }
     return ret;
 }
 
 LL ans, ret;
 int t;
 
 void solve() {
     for(ans = ; ans < phi; ++ans) {
         if(f(ans, phi) == ) {
             for(ret = ans; ret <= phi * phi; ret += phi)
                 if(f(ret, phi * phi) == ) {
                     printf("Case #%d: %d\n", t, (int)ret);
                     return ;
                 }
         }
     }
     printf("Case #%d: No solution!\n", t);
 }
 
 int main() {
     cin>>T;
     for(t = ; t <= T; ++t) {
         cin>>deg;
         for(int i = deg; i >= ; --i) cin>>a[i];
         cin>>phi;
         solve();
     }
 }