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  • HDU 5833 Zhu and 772002 （高斯消元）

    Zhu and 772002

    题目链接：

    http://acm.hdu.edu.cn/showproblem.php?pid=5833

    Description

    Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem. But 772002 has a appointment with his girl friend. So 772002 gives this problem to you. There are n numbers a1,a2,...,an. The value of the prime factors of each number does not exceed 2000, you can choose at least one number and multiply them, then you can get a number b. How many different ways of choices can make b is a perfect square number. The answer maybe too large, so you should output the answer modulo by 1000000007.

    Input

    First line is a positive integer T , represents there are T test cases. For each test case: First line includes a number n(1≤n≤300)，next line there are n numbers a1,a2,...,an,(1≤ai≤1018).

    Output

    For the i-th test case , first output Case #i: in a single line. Then output the answer of i-th test case modulo by 1000000007.

    Sample Input

    2 3 3 3 4 3 2 2 2

    Sample Output

    Case #1: 3 Case #2: 3

    Source

    2016中国大学生程序设计竞赛 - 网络选拔赛
    ##题意： 给出n个数，求有多少种方式使得选取的数的乘积是一个完全平方数.
    ##题解： 原题：[UVA11542](http://acm.hust.edu.cn/vjudge/problem/34393) (大白书P160例题25) 转化成异或方程组，并用高斯消元求解矩阵的秩. 很遗憾，上述知识点都不会....
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 2100 #define mod 1000000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); typedef long long ll; using namespace std;

    typedef int Matrix[maxn][maxn];
    int prime[maxn], vis[maxn];
    Matrix A;

    int get_primes(int m) {
    memset(vis, 0, sizeof(vis));
    int cnt = 0;
    for (int i = 2; i < m; i++) {
    if (!vis[i]) {
    prime[cnt++] = i;
    for (int j = i * i; j < m; j += i)
    vis[j] = 1;
    }
    }
    return cnt;
    }

    int gauss(Matrix A, int m, int n) {
    int i = 0, j = 0, k , r, u;
    while (i < m && j < n) {
    r = i;
    for (k = i; k < m; k++)
    if (A[k][j]) {
    r = k;
    break;
    }
    if (A[r][j]) {
    if (r != i)
    for (k = 0; k <= n; k++)
    swap(A[r][k], A[i][k]);
    for (u = i+1; u < m; u++)
    if (A[u][j])
    for (k = i; k <= n; k++)
    A[u][k] ^= A[i][k];
    i++;
    }
    j++;
    }
    return i;
    }

    LL quickmod(LL a,LL b,LL m)
    {
    LL ans = 1;
    while(b){
    if(b&1){
    ans = (ansa)%m;
    b--;
    }
    b/=2;
    a = a    a%m;
    }
    return ans;
    }

    int main() {

    //freopen("in.txt", "r", stdin);
int m = get_primes(2100);

int t;
int ca = 1;
scanf("%d", &t);
while (t--) {
    printf("Case #%d:
", ca++);
    int n, maxp = 0;;
    ll x;
    scanf("%d", &n);

    memset(A, 0, sizeof(A));
    for (int i = 0; i < n; i++) {
        scanf("%lld", &x);
        for (int j = 0; j < m; j++)
            while (x % prime[j] == 0) {
                maxp = max(maxp, j);
                x /= prime[j];
                A[j][i] ^= 1;
            }
    }

    int r = gauss(A, maxp+1, n);
    LL ans = quickmod(2, (LL)(n-r), mod) - 1;
    //printf("%lld
", (1LL << (n-r)) - 1);
    printf("%lld
", ans);
}
return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5770718.html
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