Dertouzos

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2054 Accepted Submission(s): 620

Problem Description

A positive proper divisor is a positive divisor of a number

n, excluding

n itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not.

Peter has two positive integers

n and

d. He would like to know the number of integers below

n whose maximum positive proper divisor is

Input

There are multiple test cases. The first line of input contains an integer

(1≤T≤106), indicating the number of test cases. For each test case:

The first line contains two integers

n and

(2≤n,d≤109).

Output

For each test case, output an integer denoting the answer.

Sample Input

Sample Output

Source

BestCoder Round #84

題解

題解：給你 t 個 n 和 d ，讓你求小於n的數中最大約數（不包括本身）爲d的數量。

因爲要使最大因數爲d，必存在x*d=m，同時x必須爲質數，且x必須小於d的最小質因數，因此找n前面有幾個m成立即可。因爲 t 範圍很大，所以可以先打表篩選出質數表再線性掃一遍就可以了。

之前一直超時，後來才明白判斷素數時超時

錯誤代碼

#include<iostream>
#include<string.h>
#include<queue>
#include<string>
#include<stdio.h>
#include<math.h>
using namespace std;
const int MAXN=1000010;
int prime[MAXN+1];
int getPrime(int m)
{
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=m;i++)
    {
        if(!prime[i])
            prime[++prime[0]]=i;
        for(int j=1;j<=prime[0]&&prime[j]<=m/i;j++)
        {
            prime[prime[j]*i]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,ans;
        scanf("%d%d",&n,&m);
        ans=n/m;
        if(ans>m)
        {
            getPrime(m);
            printf("%d\n",prime[0]);
        }
        else
        {
            getPrime(ans);
            printf("%d\n",prime[0]);
        }
    }
    return 0;
}

AC代碼

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
int isprime[100005]={0};
int prime[10005]={0};
int k=0;
int init()
{
    int i,j;

    for(i=2;i<100000;i++)
    {
        if(isprime[i]==0)
        {
            for(j=2;i*j<100000;j++)
            {
                isprime[i*j]=1;
            }
            prime[k]=i;//存放素數的表，
            //這k表示i（包括i）前面有幾個素數
            k++;
        }
    }
    return 0;
}
int main()
{
   int t,i;
   int n,m;
   scanf("%d",&t);
   init();
   while(t--)
   {
       scanf("%d%d",&n,&m);
       for(i=0;i<k;i++)
       {
           if(m*prime[i]>=n)
            break;
           if(m<prime[i])
            break;
           if(m%prime[i]==0)//想想爲什麼
            break;
       }
       if(m*prime[i]>=n||m<prime[i])
        i--;
        printf("%d\n",i+1);

   }
}