Project Euler_Problem 193_Few Repeated Digits_欧拉筛+容斥公式

原题目：

题目大意：

解题思路：

代码：

void serch(ll I,ll sum,ll used) {
    ll i, j, l, x,y;

    
    for (i = 1; i < I; i++) {
        if (sum * D[i] > N)break;
        x = sum * D[i];
        y = N / x;

        if (used % 2 == 0) {
            ans1 = ans1 - y;
        }
        else {
            ans1 = ans1 + y;
        }

        serch(i, x, used + 1);

    }

}


void solve() {
    ll i, j,k,x,y,z,p,q,u,v;
    N = (( (ll)1 )<<50), NN = 4;
    //N = 20;
    double a, b, c,d;
    M.NT.get_prime_Euler(100000000);
    
    for (i = 1; i <= M.NT.pcnt; i++) {
        x = M.NT.prime[i];
        if (x * x > N)break;
    }

    ll len = i - 1;

    for (i = 1; i <= len; i++) {
        D[i] = M.NT.prime[i] * M.NT.prime[i];
        ans1 = ans1 + N / D[i];
    }

    for (i = len; i >= 1; i--) {
        serch(i, D[i], 2);
    }
    

    printf("%lld\n",N-ans1);
}