SGU 200. Cracking RSA(高斯消元+高精度)

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題目大意：給出m個整數，因子全部存在於前t個素數。問有多少個子集，他們的乘積是平方數。

解題思路：

完全平方數就是要求每個質因子的指數是偶數次。
對每個質因子建立一個方程。 變成模2的線性方程組。
求解這個方程組有多少個自由變元，答案就是 2^p - 1 。(-1是去掉空集的情況)

注意由於2^p會超出資料範圍所以還需要用高精度演算法。

200. Cracking RSAtime limit per test: 0.25 sec.
memory limit per test: 65536 KBinput: 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;}