hdu 5750——Dertouzos（简单）

Problem Description
A positive proper divisor is a positive divisor of a number nn, excluding nn itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not.

Peter has two positive integers nn and dd. He would like to know the number of integers below nn whose maximum positive proper divisor is dd.

题目大意：正整数x称为n的positive proper divisor, 当且仅当x | n并且1≤ x 且n＞x。 例如, 1, 2, 和3是6的positive proper divisor, 但是6不是
Peter给你两个正整数n和d. 他想要知道有多少小于n的整数, 满足他们的最大positive proper divisor恰好是d.

题解：对于每个整数，它乘以的数必须都小于等于它的最小因数，这样才能保证其满足条件。

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>using namespace std;const int maxn = 100000 + 130;const int maxm = 1e5+120;int prime[maxn];bool vis[maxm]; int mind[maxn];int n,d,cnt;void get_prime(int n){    memset(vis,0,sizeof(vis));    memset(mind,0,sizeof(mind));    vis[1]=1;cnt=0;    for(int i=2;i * i<n;i++){        if(vis[i]) continue;        else mind[i] = i;        for(int j=i*i;j<n;j+=i){            vis[j]= 1;            if(!mind[j]) mind[j]=i;            }    }    for(int i=2;i<n;i++) if(!vis[i]) prime[++cnt]=i;}int main(){    int kase;scanf("%d",&kase);    get_prime(maxm-10);    while(kase--){        scanf("%d%d",&n,&d);        if(d>=n) {printf("0\n");continue;}        int ans = 0 ;        int maxnum = min(d,(n-1)/d);        for(int i=1;i<=cnt;i++){            if(prime[i]*d>=n) break;            ans++;            if(d%prime[i]==0) break;        }        printf("%d\n",ans);    }    return 0;}