HDU5750 Dertouzos

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 d.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:

The first line contains two integers n and d (2≤n,d≤109).

 

Output

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

 

Sample Input

910 210 310 410 510 610 710 810 9100 13

 

Sample Output

121000004

 

Source

BestCoder Round #84

题解

题解：给你 t 个 n 和 d ，让你求小于n的数中最大约数（不包括本身）为d的数量。   

       因为要使最大因数为d，必存在x*d=m，同时x必须为质数，且x必须小于d的最小质因数，因此找n前面有几个m成立即可。因为 t 范围很大，所以可以先打表筛选出质数表再线性扫一遍就可以了。

之前一直超时，后来才明白判断素数时超时

错误代码

#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代码

#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;//存放素数的表，            //这k表示i（包括i）前面有几个素数            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)//想想为什么            break;       }       if(m*prime[i]>=n||m<prime[i])        i--;        printf("%d\n",i+1);   }}