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)
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
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;
}