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);
}