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SGU 200. Cracking RSA(高斯消元+高精度)

2024-07-14 22:16:01   221人阅读

题目大意：给出m个整数，因子全部存在于前t个素数。问有多少个子集，他们的乘积是平方数。

解题思路：

完全平方数就是要求每个质因子的指数是偶数次。
对每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自由变元，答案就是 2^p - 1 。(-1是去掉空集的情况)

注意由于2^p会超出数据范围所以还需要用高精度算法。

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, ..., bm 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 bi 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 bi 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 ≤ bi ≤ 109 for all i. 

Output

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


Sample test(s)

Input
3 4 9 20 500 3 
Output

3 
[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;
}
































        

            
                
                  
                
                

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