HDU-4569 Special equations （利用数学知识巧妙降低复杂度）

最新推荐文章于 2017-08-04 15:21:46 发布

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最新推荐文章于 2017-08-04 15:21:46 发布

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本文介绍了一个解决特定形式的同余方程f(x) ≡ 0 (mod p^2)的方法，其中p为质数。输入包括多项式的系数和质数p，输出为满足条件的一个解或无解提示。

　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!

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long LL;
LL a[5];
LL f(int n,LL x)
{
    LL sum=0;
    for(int i=1;i<=n;i++)
    {
        sum+=a[i]*x;
        x*=x;
    }
    sum+=a[0];
    return sum;
}
void solve(int n,LL pri)
{
    for(LL i=0;i<pri;i++)
        if(f(n,i)%pri==0)
            for(LL j=i;j<=pri*pri;j+=pri)
                if(f(n,j)%(pri*pri)==0)
        {
            printf("%lld\n",j);
            return;
        }
    printf("No solution!\n");
}
int main()
{
    int t,n;
    LL pri;
    cin>>t;
    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d",&n);
        for(int i=n;i>=0;i--)
            scanf("%lld",&a[i]);
        scanf("%lld",&pri);
        printf("Case #%d: ",ca);
        solve(n,pri);
    }
    return 0;
}