HDU 4569 Special equations(取模)

Special equations

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4569

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!"

 

Sample Input

4

2 1 1 -5 7

1 5 -2995 9929

2 1 -96255532 8930 9811

4 14 5458 7754 4946 -2210 9601

 

Sample Output

Case #1: No solution!
Case #2: 599
Case #3: 96255626
Case #4: No solution!

 

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题意：给你函数 f(x)

_{最多N就4位，输入任意一个x使f(x)%(prime*prime)=0。}

首先要找一个 f(x) % prime = 0 ，只有满足这个条件 f(x) % (prime * prime) 才有可能等于0，对于f(x) % prime，可以枚举 0 - prime，如果满足条件那么在判断（ x + prime * k ) % （prime * prime)是否满足条件

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 typedef long long LL;
 LL a[], P;
 int deg;
 bool func(LL x, LL mod)
 {
     LL res = , now = ;
     for (int i = ; i <= deg; i++)
     {
         res = (res + now * a[i]) % mod;
         now = (now * x) % mod;
     }
     if (res % mod)
         return ;
     return ;
 }
 int main()
 {
     int test;
     scanf("%d", &test);
     for (int t = ; t <= test; t++)
     {
         scanf("%d", &deg);
         for (int i = deg; i >= ; i--)
         {
             scanf("%I64d", &a[i]);
         }
         scanf("%I64d", &P);
         LL ans = -;
         int flag = false;
         for (LL i = ; i <= P; i++)
         {
             if (func(i, P))
                 continue;
             for (LL j = i; j <= P * P; j += P)
             {
                 if (func(j, P * P))
                     continue;
                 flag = true;
                 ans = j;
                 break;
             }
             if (flag)
                 break;
         }
         printf("Case #%d: ", t);
         if (flag)
             printf("%I64d\n", ans);
         else
             printf("No solution!\n");
     }
     return ;
 }