hdu4407容斥原理

本文主要是介绍hdu4407容斥原理，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题意：
有一个元素为 1~n 的数列{An}，有2种操作（1000次）：
1、求某段区间 [a,b] 中与 p 互质的数的和。
2、将数列中某个位置元素的值改变。

import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.NavigableSet;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.SortedSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;public class Main {public static void main(String[] args) throws IOException{StreamTokenizer cin = new StreamTokenizer(new BufferedInputStream(System.in)); InputReader in = new InputReader(System.in)  ;PrintWriter out = new PrintWriter(System.out) ;int t = in.nextInt() ;for(int i = 1 ; i <= t ; i++){new Task().solve(in, out)  ;  // out.flush() ;}out.flush() ; }}class  Task{static int N = 700 ;static int[] prime = new int[N] ;static int pid = 0 ;static boolean[] vis = new boolean[N] ;static{for(int i = 2 ; i < N ; i++){if(! vis[i]) prime[pid++] = i ;for(int j = 0 ; j < pid && prime[j] * i < N ; j++){vis[i * prime[j]] = true ;if(i % prime[j] == 0) break ;}}}int  gcd(int x , int y){return y == 0 ? x : gcd(y , x%y) ; }HashMap<Integer, Integer> hash = new HashMap<Integer, Integer>() ;ArrayList<Integer> factor = new ArrayList<Integer>() ;long  sigma(long n){return (n + 1 ) * n / 2 ;}long  sum(int n){long ans = sigma(n) ;int m = factor.size() ;int limit = 1<<m ;for(int i = 1 ; i < limit ; i++){int k = 0 ;int val = 1 ;for(int j = 0 ; j < m ; j++){if((i & (1<<j)) > 0){k++ ;val *= factor.get(j)  ;}}long t = sigma(n/val) * val ;if((k&1) > 0) ans -= t ;else ans += t ;}return ans ;}long  ask(int left , int right , int p){    int n = p ;factor.clear();for(int i = 0 ; i < pid && prime[i]*prime[i] <= n ; i++){if(n % prime[i] == 0){while(n % prime[i] == 0) n /= prime[i] ;factor.add(prime[i]) ;}}if(n != 1) factor.add(n) ;long ans = sum(right) - sum(left-1) ;for(Map.Entry<Integer, Integer> e : hash.entrySet()){int id = e.getKey() ;int val = e.getValue() ;if(left <= id && id <= right){if(gcd(id , p) == 1) ans -= id ;if(gcd(val ,p) == 1) ans += val ;}}return ans ;}public void solve(InputReader in , PrintWriter out) throws IOException{int n = in.nextInt() ;int m = in.nextInt() ;while(m-- > 0){if(in.nextInt() == 1) out.println(ask(in.nextInt() , in.nextInt() , in.nextInt())) ;else  hash.put(in.nextInt(), in.nextInt()) ;}}
}class   InputReader{public BufferedReader  reader;public StringTokenizer  tokenizer;public InputReader(InputStream stream){reader = new BufferedReader(new InputStreamReader(stream), 32768) ;tokenizer = null ;}public String next(){while(tokenizer == null || ! tokenizer.hasMoreTokens()){try{tokenizer = new StringTokenizer(reader.readLine());}catch (IOException e){throw new RuntimeException(e);}}return tokenizer.nextToken();  }public int  nextInt(){return Integer.parseInt(next());}public long nextLong(){return Long.parseLong(next());}public double nextDouble(){return  Double.parseDouble(next());}}

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