题目链接：http://codeforces.com/problemset/problem/546/D
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 }
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);
}
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
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);
}
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();
}
}
时间： 2024-10-22 10:24:13