Special equations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 206 Accepted Submission(s): 108
Special Judge

Problem Description
　　Let f(x) = anxn +...+ a1x +a0, in which ai (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 an to a0 (0 < abs(an) <= 100; abs(ai) <= 10000 when deg >= 3, otherwise abs(ai) <= 100000000, i 　　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!

Source
2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

题解：f(x)%(p*p)=0那么一定有f(x)%p=0，f(x)%p=0那么一定有f(x+p)%p=0。

所以我们可以开始从0到p枚举x，当f(x)%p=0，然后再从x到p*p枚举，不过每次都是+p，找到了输出即可，没有的话No solution!，这里只需要枚举到pri*pri，这个证法和前面的一样

#include   
  
int a[6],pri,n;  
  
__int64 getf(int x,int i)  
{  
    int j;  
    __int64 sum=0;  
    for(j=0;j

int a[6],pri,n;

__int64 getf(int x,int i)
{
	int j;
	__int64 sum=0;
	for(j=0;j