HDU 4569 (推导)

最新推荐文章于 2021-07-08 19:01:28 发布

morejarphone

最新推荐文章于 2021-07-08 19:01:28 发布

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分类专栏：数论/推导

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本文介绍了一种解决特殊方程模平方素数为0的问题的方法。通过先找到方程模素数为0的解，再验证这些解加上素数倍数是否满足原方程模平方素数为0。

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Special equations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 435 Accepted Submission(s): 274
Special Judge

Problem Description

　　Let f(x) = a_nxⁿ +...+ a₁x +a₀, 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 a₀ (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!

题意：求一个方程模m^2为0是否有解。

因为m是素数，所以方程模m^2为0必然需要方程模m为0，而所有的x(x>=m)模m为0，必

然有(x-m)模m为0.

所以就可以寻找[0,m-1]中f(x)模m为0的x，然后判断x+m,x+2m....是不是满足。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <map>
#include <vector>
#include <math.h>
#include <queue>
using namespace std;
#define maxn 11

int n;
long long a[maxn];
long long m, mm;

long long f1 (long long x) {
    long long ans = 0;
    for (int i = n; i >= 0; i--) {
        long long cur = 1;
        for (int j = 0; j < i; j++) cur *= x, cur %= m;
        ans += cur*a[i];
        ans = (ans%m+m)%m;
    } 
    return ans;
}

long long f2 (long long x) {
    long long ans = 0;
    for (int i = n; i >= 0; i--) {
        long long cur = 1;
        for (int j = 0; j < i; j++) cur *= x, cur %= mm;
        ans += cur*a[i];
        ans = (ans%mm+mm)%mm;
    } 
    return ans;
}

int main () {
    //freopen ("in.txt", "r", stdin);
    int t, kase = 0;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        for (int i = n; i >= 0; i--) {
            scanf ("%lld", &a[i]);
        }
        scanf ("%lld", &m);
        mm = m*m;
        printf ("Case #%d: ", ++kase);
        long long x, ans;
        for (x = 0; x < m; x++) {
            if (f1 (x) == 0) {
                for (ans = x; ans < m*m; ans += m) {
                    if (f2 (ans) == 0) {
                        printf ("%lld\n", ans);
                        goto out;
                    }
                }
            }
        }
        printf ("No solution!\n");
        out: ;
    }
    return 0;
}