Codeforces D546：Soldier and Number Game

题目链接

输入t对数 a， b

求（b,a]内的每个数拆成素因子的个数和

这里每个数都可以写成素数的乘积，可以写成几个素数的和就有几个素因子，这里求的是（b,a]内的素因子和

思路：

素数的素因子个数是1

对于非素数A的素因子个数 = A/k + 1 其中k是素数，也是第一个素数，或者K是比A小的数，并且A%k==0

下面是利用K是比A小的数，并且A%k==0

 1     void solve(){
 2         Scanner sc = new Scanner(System.in);
 3         int t = sc.nextInt();
 4         int a ,b;
 5         int MAX = 5000000;
 6         int[] arr=new int[MAX+1];
 7         for(int i=2;i<=MAX;i++){
 8             if(i%2==0){
 9                 arr[i] = arr[i/2]+1;
10                 continue;
11             }
12             for(int j=3;j*j<=i;j+=2){
13                 int k=i/j;
14                 if(k*j==i){
15                     arr[i] = arr[k] + 1;
16                     break;
17                 }
18             }
19             if (arr[i]==0)
20                 arr[i] = 1;
21         }
22         for(int i=2;i<=MAX;i++)
23             arr[i]+=arr[i-1];
24         while(t!=0){
25         a = sc.nextInt();
26         b = sc.nextInt();
27         System.out.println(arr[a]-arr[b]);
28         t--;
29         }
30     }

View Code

这个读数据效率低，造成运行时间超时

这个是利用素数的，找到第一个素数K，并且A%k==0

    void solve3() throws NumberFormatException, IOException{int limit = 5000005;BufferedReader br=new BufferedReader(new InputStreamReader(System.in));int t=Integer.parseInt(br.readLine());StringTokenizer st;int div[] = new int[limit];boolean isPrime[] = new boolean[limit];int prime[] = new int[limit];int p=0;for(int i=2;i<limit;i++){if(!isPrime[i])prime[p++]=i;for(int j=0;j<p &&i*prime[j]<limit;j++){isPrime[i*prime[j]] = true;if(i%prime[j]==0)break;}}for(int i=2;i<limit;i++){int k = i;if(!isPrime[i]){div[i] = 1;continue;}for(int j=0;j<p;j++){if(k%prime[j]==0){div[i] = div[k/prime[j]] + 1;break;}}}for(int i=2;i<limit;i++)div[i] += div[i-1];StringBuilder sb=new StringBuilder();for(int i=0;i<t;i++){st=new StringTokenizer(br.readLine());int a=Integer.parseInt(st.nextToken());int b=Integer.parseInt(st.nextToken());sb.append(div[a]-div[b]);sb.append('\n');}System.out.print(sb);}

View Code

换了读数据的方式，成功通过，这里利用到，素数筛选法求出素数的集合

参考链接

还有下面一种，都是参考素数筛选法的求解

void solve2() throws NumberFormatException, IOException{BufferedReader br=new BufferedReader(new InputStreamReader(System.in));int t=Integer.parseInt(br.readLine());StringTokenizer st;int limit = 5000000;boolean[] array = new boolean[limit+1];int div[] = new int[limit+1];for(int i=2;i<=limit;i++){if(array[i]) continue;for(int j=i;j<=limit;j+=i){array[j] = true;int k=j;int temp=0;while(k%i==0){k/=i;temp++;}div[j]+=temp;}}for(int i=1;i<=limit;i++)div[i]+=div[i-1];StringBuilder sb=new StringBuilder();for(int i=0;i<t;i++){st=new StringTokenizer(br.readLine());int a=Integer.parseInt(st.nextToken());int b=Integer.parseInt(st.nextToken());sb.append(div[a]-div[b]);sb.append('\n');}System.out.print(sb);}

View Code

全部程序如下

package codeforces;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class D546 {void run() throws NumberFormatException, IOException{
//        solve();// 超时
        solve3();}void solve3() throws NumberFormatException, IOException{int limit = 5000005;BufferedReader br=new BufferedReader(new InputStreamReader(System.in));int t=Integer.parseInt(br.readLine());StringTokenizer st;int div[] = new int[limit];boolean isPrime[] = new boolean[limit];int prime[] = new int[limit];int p=0;for(int i=2;i<limit;i++){if(!isPrime[i])prime[p++]=i;for(int j=0;j<p &&i*prime[j]<limit;j++){isPrime[i*prime[j]] = true;if(i%prime[j]==0)break;}}for(int i=2;i<limit;i++){int k = i;if(!isPrime[i]){div[i] = 1;continue;}for(int j=0;j<p;j++){if(k%prime[j]==0){div[i] = div[k/prime[j]] + 1;break;}}}for(int i=2;i<limit;i++)div[i] += div[i-1];StringBuilder sb=new StringBuilder();for(int i=0;i<t;i++){st=new StringTokenizer(br.readLine());int a=Integer.parseInt(st.nextToken());int b=Integer.parseInt(st.nextToken());sb.append(div[a]-div[b]);sb.append('\n');}System.out.print(sb);}void solve2() throws NumberFormatException, IOException{BufferedReader br=new BufferedReader(new InputStreamReader(System.in));int t=Integer.parseInt(br.readLine());StringTokenizer st;int limit = 5000000;boolean[] array = new boolean[limit+1];int div[] = new int[limit+1];for(int i=2;i<=limit;i++){if(array[i]) continue;for(int j=i;j<=limit;j+=i){array[j] = true;int k=j;int temp=0;while(k%i==0){k/=i;temp++;}div[j]+=temp;}}for(int i=1;i<=limit;i++)div[i]+=div[i-1];StringBuilder sb=new StringBuilder();for(int i=0;i<t;i++){st=new StringTokenizer(br.readLine());int a=Integer.parseInt(st.nextToken());int b=Integer.parseInt(st.nextToken());sb.append(div[a]-div[b]);sb.append('\n');}System.out.print(sb);}void solve(){Scanner sc = new Scanner(System.in);int t = sc.nextInt();int a ,b;int MAX = 5000000;int[] arr=new int[MAX+1];for(int i=2;i<=MAX;i++){if(i%2==0){arr[i] = arr[i/2]+1;continue;}for(int j=3;j*j<=i;j+=2){int k=i/j;if(k*j==i){arr[i] = arr[k] + 1;break;}}if (arr[i]==0)arr[i] = 1;}for(int i=2;i<=MAX;i++)arr[i]+=arr[i-1];while(t!=0){        a = sc.nextInt();b = sc.nextInt();System.out.println(arr[a]-arr[b]);t--;}}public static void main(String[] args) throws NumberFormatException, IOException {new D546().run();}}

View Code

转载于:https://www.cnblogs.com/theskulls/p/4851288.html

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