容斥原理 數論

兩個集合的容斥關係公式：A∪B =|A∪B| = |A|+|B| - |A∩B |(∩：重合的部分）

三個集合的容斥關係公式：|A∪B∪C| = |A|+|B|+|C| - |A∩B| - |B∩C| - |C∩A| + |A∩B∩C|

最後可以推廣到n個集合，集合裏的元素爲奇數則加，偶數減

百度有證明和詳細說明

http://baike.baidu.com/link?url=ONESj-EzjxEKYUdznssG9xuOU9bGQWGi0JdxuzGnW6FqGipdVvhLjjX_n52gEg2mJaBNGY7Nf6GlEfiquqbhY_

hdu 4135

很簡單，直接求出所有的質因子，然後容斥解決

/********************************************
Author         :Crystal
Created Time   :
File Name      :
********************************************/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <climits>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <sstream>
#include <cctype>
using namespace std;
typedef long long ll;
typedef pair<int ,int> pii;
#define MEM(a,b) memset(a,b,sizeof a)
#define CLR(a) memset(a,0,sizeof a);
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
//#define LOCAL
std::vector<int> v;
void gao(int n){
	v.clear();
	int res = n;
	for(int i=2;i*i<=res;i++){
		if(res%i==0){
			v.push_back(i);
			while(res%i==0)res/=i;
		}
	}
	if(res > 1)v.push_back(res);
}
ll cal(ll n){
	ll sum = 0;
	sum = n;
	int s = v.size();
	for(int i=1;i<=(1<<s)-1;i++){
		int cnt = 0;
		ll ans = 1;
		for(int j=0;j<s;j++){
			if(i & (1<<j)){
				cnt++;
				ans *= v[j];
			}
		}
		if(cnt%2){
			sum -= n/ans;
		}
		else sum += n/ans;
	}
	return sum;
}
int main()
{
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	int t;cin >> t;
	int kase = 1;
	while(t--){
		ll a,b,n;
		cin >> a >> b >> n;
		gao(n);
		printf("Case #%d: %lld\n",kase++,cal(b)-cal(a-1));
	}
	return 0;
}