SGU 200. Cracking RSA(高斯消元+高精度)

2021-09-18 20:29:53

标题效果：鉴于m整数，之前存在的所有因素t素数。问：有多少子集。他们的产品是数量的平方。

解题思路：

全然平方数就是要求每一个质因子的指数是偶数次。

对每一个质因子建立一个方程。

变成模2的线性方程组。

求解这个方程组有多少个*变元。答案就是 2^p - 1 。(-1是去掉空集的情况)

注意因为2^p会超出数据范围所以还须要用高精度算法。

200. Cracking RSA

time limit per test: 0.25 sec.

memory limit per test: 65536 KB

input: standard

output: standard

The following problem is somehow related to the final stage of many famous integer factorization algorithms involved in some cryptoanalytical problems, for example cracking well-known RSA public key system.

The most powerful of such algorithms, so called quadratic sieve descendant algorithms, utilize the fact that if n = pq where p and q are large unknown primes needed to be found out, then if v²=w²(mod n), u ≠ v (mod n) and u ≠ -v (mod n),
then gcd(v + w, n) is a factor of n (either p or q).

Not getting further in the details of these algorithms, let us consider our problem. Given m integer numbers b₁, b₂, ..., b_m such that all their prime factors are from the set of first t primes, the task is to find such a subset
S of {1, 2, ..., m} that product of b_i for i from S is a perfect square i.e. equal to u² for some integer u. Given such S we get one pair for testing (product of S elements stands for v when w is known from other steps of algorithms which
are of no interest to us, testing performed is checking whether pair is nontrivial, i.e. u ≠ v (mod n) and u ≠ -v (mod n)). Since we want to factor n with maximum possible probability, we would like to get as many such sets as possible. So the interesting
problem could be to calculate the number of all such sets. This is exactly your task.

Input

The first line of the input file contains two integers t and m (1 ≤ t ≤ 100, 1 ≤ m ≤ 100). The second line of the input file contains m integer numbers b_i such that all their prime factors are from t first primes (for example, if t = 3 all their
prime factors are from the set {2, 3, 5}). 1 ≤ b_i ≤ 10⁹ for all i.

Output

Output the number of non-empty subsets of the given set {b_i}, the product of numbers from which is a perfect square

Sample test(s)

Input

3 4 9 20 500 3

Output

problem=200" style="color:rgb(34,68,136); text-decoration:none">[submit]

#include <algorithm>

#include <iostream>

#include <stdlib.h>

#include <string.h>

#include <iomanip>

#include <stdio.h>

#include <string>

#include <queue>

#include <cmath>

#include <stack>

#include <map>

#include <set>

#define eps 1e-10

///#define M 1000100

#define LL __int64

///#define LL long long

///#define INF 0x7ffffff

#define INF 0x3f3f3f3f

#define PI 3.1415926535898

#define zero(x) ((fabs(x)<eps)?0:x)

const int maxn = 210;

using namespace std;

bool f[maxn+1000];

int k[maxn+1000];

int a[maxn][maxn];

int num[maxn];

int equ, var;

char str1[maxn], str2[maxn];

void Add(char a[], char b[], char c[])

{

    int len1 = strlen(a);

    int len2 = strlen(b);

    int n = max(len1, len2);

    int add = 0;

    for(int i = 0; i < n; i++)

    {

        int cnt = 0;

        if(i < len1) cnt += a[i]-'0';

        if(i < len2) cnt += b[i]-'0';

        cnt += add;

        add = cnt/10;

        c[i] = cnt%10+'0';

    }

    if(add) c[n++] = add+'0';

    c[n] = 0;

}

void Sub_1(char a[])

{

    int s = 0;

    while(a[s] == '0') s++;

    a[s]--;

    for(int i = 0; i < s; i++)

        a[i] = '9';

    int len = strlen(a);

    while(len > 1 && a[len-1] == '0') len--;

    a[len] = 0;

}

void Prime()

{

    int t = 0;

    memset(f, false, sizeof(f));

    for(int i = 2; i <= 1005; i++)

    {

        if(!f[i])

            k[t++] = i;

        for(int j = 0; j < t; j++)

        {

            if(i*k[j] > 1005)

                break;

            f[i*k[j]] = true;

            if(i%k[j] == 0)

                break;

        }

    }

}

int Gauss()

{

    int row, col;

    int max_r;

    row = col = 0;

    while(row < equ && col < var)

    {

        max_r = row;

        for(int i = row+1; i < equ; i++)

        {

            if(a[i][col]) max_r = i;

        }

        if(a[max_r][col] == 0)

        {

            col++;

            continue;

        }

        if(max_r != row)

        {

            for(int j = col; j <= var; j++) swap(a[max_r][j], a[row][j]);

        }

        for(int i = row+1; i < equ; i++)

        {

            if(a[i][col] == 0) continue;

            for(int j = col; j <= var; j++) a[i][j] ^= a[row][j];

        }

        col++;

        row++;

    }

    return var-row;

}

int main()

{

    Prime();

    int n, m;

    while(cin >>n>>m)

    {

        memset(a, 0, sizeof(a));

        for(int i = 0; i < m; i++) cin >>num[i];

        equ = n;

        var = m;

        for(int i = 0; i < n; i++)

        {

            for(int j = 0; j < m; j++)

            {

                int ans = 0;

                while(num[j]%k[i] == 0)

                {

                    ans ++;

                    num[j]/=k[i];

                }

                if(ans%2) a[i][j] = 1;

            }

        }

        int N = Gauss();

        strcpy(str1, "1");

        for(int i = 0; i < N; i++)

        {

            Add(str1, str1, str2);

            strcpy(str1, str2);

        }

        Sub_1(str1);

        for(int i = strlen(str1)-1; i >= 0; i--) cout<<str1[i];

        cout<<endl;

    }

    return 0;

}

码农公寓

200. Cracking RSA

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