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题目链接:
Time Limit: 7000/3500 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-8;
int prime[N],sum[N],a[N],cnt=0,n,d;
void Init()
{
sum[1]=0;
For(i,2,N-maxn)
{
if(!prime[i])
{
for(int j=2*i;j<N-maxn;j+=i)
{
prime[j]=1;
}
sum[i]=sum[i-1]+1;
}
else sum[i]=sum[i-1];
}
For(i,2,N-maxn)
{
if(!prime[i])a[++cnt]=i;
}
}
inline int check(int x)
{
for(int i=1;i<=cnt;i++)
{
if(x%a[i]==0)return a[i];
if((LL)a[i]*a[i]>x||a[i]>n/d)break;
}
return x;
}
int main()
{
int t;
read(t);
Init();
while(t--)
{
read(n);read(d);
n--;
int le=min(check(d),n/d);
printf("%d\n",sum[le]);
}
return 0;
}
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5699905.html