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 _nx ⁿ +...+ a ₁x +a ₀, 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 a ₀ (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）：

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <cstring>

 4 #include <queue>

 5 using namespace std;

 6 typedef long long LL;

 7 

 8 const int MAXN = 8;

 9 

10 int T, deg;

11 LL phi;

12 LL a[MAXN];

13 

14 LL f(LL x, LL m) {

15     LL ret = 0, xx = 1;

16     for(int i = 0; i <= deg; ++i) {

17         ret = (ret + a[i] * xx) % m;

18         xx = (xx * x) % m;

19     }

20     return ret;

21 }

22 

23 LL ans, ret;

24 int t;

25 

26 void solve() {

27     for(ans = 0; ans < phi; ++ans) {

28         if(f(ans, phi) == 0) {

29             for(ret = ans; ret <= phi * phi; ret += phi)

30                 if(f(ret, phi * phi) == 0) {

31                     printf("Case #%d: %d\n", t, (int)ret);

32                     return ;

33                 }

34         }

35     }

36     printf("Case #%d: No solution!\n", t);

37 }

38 

39 int main() {

40     cin>>T;

41     for(t = 1; t <= T; ++t) {

42         cin>>deg;

43         for(int i = deg; i >= 0; --i) cin>>a[i];

44         cin>>phi;

45         solve();

46     }

47 }

View Code