hdu4589 Special equations(数论)

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) = 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!

 

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<stdio.h>

int a[6],pri,n;

__int64 getf(int x,int i)
{
	int j;
	__int64 sum=0;
	for(j=0;j<i;j++)
	{
		sum=(sum+a[j])*x;
	}
	return sum+a[j];
}

void solve()
{
	int i,j,k;
	int pri2=pri*pri;
	for(i=0;i<pri;i++)
	{
		if((getf(i,n))%pri==0)
		{
			for(j=i;j<pri2;j+=pri)
			{
				if(getf(j,n)%pri2==0)
				{
					printf("%d
",j);
					return ;
				}
			}
		}
	}
	printf("No solution!
");
}
int main()
{
	int i,j,k,t,no;
	__int64 temp;
	scanf("%d",&t);
	for(k=1;k<=t;k++)
	{
		scanf("%d",&n);
		for(i=0;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		scanf("%d",&pri);
		printf("Case #%d: ",k);
		solve();
	}
	return 0;
}