容斥原理学习(Hdu 4135，Hdu 1796）

题目链接Hdu4135

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1412    Accepted Submission(s): 531

Problem Description

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

 

Input

The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10¹⁵) and (1 <=N <= 10⁹).

 

Output

For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

 

Sample Input

2
1 10 2
3 15 5

 

Sample Output

Case #1: 5
Case #2: 10

Hint

In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

题意：求区间[a, b]内与ｎ互质的数的个数。

思路：如果某个数与ｎ互质，那么这个数一定和ｎ没有公共因子。所以题目就转化为有多少个数与ｎ无公共因子。

可以通过求区间内有多少个数与ｎ存在公共因子来得到答案。筛去２的倍数，３的倍数，５的倍数。。。容斥就可以啦。

Accepted Code:

 /*************************************************************************
     > File Name: 4135.c
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月05日 星期五 16时33分45秒
     > Propose:
  ************************************************************************/
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 using namespace std;
 /*Let's fight!!!*/

 typedef long long LL;

 LL gcd(LL a, LL b) {
     if (!b) return a;
     return gcd(b, a % b);
 }

 LL cal(const vector<int> &var, const LL &n) {
       LL sz = var.size(), res = ;
       for (LL i = ; i < (<<sz); i++) {
           int num = ;
           for (LL j = i; j != ; j >>= ) if (j & ) num++;
           LL lcm = ;
           for (LL j = ; j < sz; j++) {
               if ((i >> j) & ) lcm = lcm / gcd(lcm, var[j]) * var[j];
               if (lcm > n) break;
           }
           if (num %  == ) res -= n / lcm;
           else res += n / lcm;
       }

       return res;
 }

 int main(void) {
     ios::sync_with_stdio(false);
     int T, cas = ;
     cin >> T;
     while (T--) {
         LL a, b, n, x;
         cin >> a >> b >> n;

         vector<int> var;
         x = n;
         for (int i = ; i * i <= x; i++) {
               if (x % i == ) {
                 var.push_back(i);
                 while (x % i == ) x /= i;
             }
         }
         if (x > ) var.push_back(x);

         LL res = b - a +  - cal(var, b) + cal(var, a - );
         cout << "Case #" << cas++ << ": " << res << endl;
     }

     return ;
 }

题目链接Hdu1796

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4205    Accepted Submission(s): 1198

Problem Description

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

Output

  For each case, output the number.

Sample Input

12 2
2 3

Sample Output

7

和上题一样。。。

Accepted Code:

/*************************************************************************
    > File Name: 1796_dfs.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年09月06日 星期六 08时28分01秒
    > Propose:
 ************************************************************************/

#include <cmath>
#include <string>
#include <cstdio>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
/*Let's fight!!!*/

typedef long long LL;
int a[], n, m;

LL gcd(LL a, LL b) {
      if (!b) return a;
    return gcd(b, a % b);
}

void dfs(LL now, int num, LL lcm, LL &res) {
      lcm = lcm / gcd(lcm, a[now]) * a[now];
    if (num %  == ) res -= n / lcm;
    else res += n / lcm;
    for (int i = now + ; i < m; i++)
          dfs(i, num + , lcm, res);
}

int main(void) {
      ios::sync_with_stdio(false);
    while (cin >> n >> m) {
          int cnt = ;
        for (int i = ; i < m; i++) {
              int x;
            cin >> x;
            if (x > ) a[cnt++] = x;
        }
        m = cnt;
        LL res = ;
        n--;
        for (int i = ; i < m; i++) {
              dfs(i, , a[i], res);
        }

        cout << res << endl;
    }
}

//位运算实现
/*************************************************************************
    > File Name: 1796.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年09月05日 星期五 21时32分48秒
    > Propose:
 ************************************************************************/

#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
/*Let's fight!!!*/

typedef long long LL;
int a[];

LL gcd(LL a, LL b) {
    if (!b) return a;
    return gcd(b, a % b);
}

int main(void) {
    ios::sync_with_stdio(false);
    LL n, m;
    while (cin >> n >> m) {
        int d = ;
        for (int i = ; i < m; i++) {
            int x;
            cin >> x;
            if (x >  && x <= n) a[d++] = x;
        }

        n--;
        LL res = ;
        for (LL i = ; i < ( << d); i++) {
              int num = ;
            for (LL j = i; j != ; j >>= ) num += j & ;
            LL lcm = ;
            for (LL j = ; j < d; j++) {
                  if ((i >> j) & ) {
                    lcm = lcm / gcd(lcm, a[j]) * a[j];
                    if (lcm > n) break;
                }
            }
            if (num %  == ) res -= n / lcm;
            else res += n / lcm;
        }

        cout << res << endl;
    }
    return ;
}