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容斥原理学习(Hdu 4135,Hdu 1796)

时间:2014-09-06 09:39:43      阅读:320      评论:0      收藏:0      [点我收藏+]

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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:

 1 /*************************************************************************
 2     > File Name: 4135.c
 3     > Author: Stomach_ache
 4     > Mail: sudaweitong@gmail.com
 5     > Created Time: 2014年09月05日 星期五 16时33分45秒
 6     > Propose: 
 7  ************************************************************************/
 8 #include <cstdio>
 9 #include <vector>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 using namespace std;
14 /*Let‘s fight!!!*/
15 
16 typedef long long LL;
17 
18 LL gcd(LL a, LL b) {
19     if (!b) return a;
20     return gcd(b, a % b);
21 }
22 
23 LL cal(const vector<int> &var, const LL &n) {
24       LL sz = var.size(), res = 0;
25       for (LL i = 1; i < (1<<sz); i++) {
26           int num = 0;
27           for (LL j = i; j != 0; j >>= 1) if (j & 1) num++;
28           LL lcm = 1;
29           for (LL j = 0; j < sz; j++) {
30               if ((i >> j) & 1) lcm = lcm / gcd(lcm, var[j]) * var[j];
31               if (lcm > n) break;
32           }
33           if (num % 2 == 0) res -= n / lcm; 
34           else res += n / lcm;
35       }
36 
37       return res;
38 }
39 
40 int main(void) {
41     ios::sync_with_stdio(false);
42     int T, cas = 1;
43     cin >> T;
44     while (T--) {
45         LL a, b, n, x;
46         cin >> a >> b >> n;
47 
48         vector<int> var;
49         x = n;
50         for (int i = 2; i * i <= x; i++) {
51               if (x % i == 0) {
52                 var.push_back(i);
53                 while (x % i == 0) x /= i;
54             }
55         }
56         if (x > 1) var.push_back(x);
57 
58         LL res = b - a + 1 - cal(var, b) + cal(var, a - 1);
59         cout << "Case #" << cas++ << ": " << res << endl;
60     }
61 
62     return 0;
63 }

 

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[15], 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 % 2 == 0) res -= n / lcm;
    else res += n / lcm;
    for (int i = now + 1; i < m; i++) 
          dfs(i, num + 1, lcm, res);
}

int main(void) {
      ios::sync_with_stdio(false);
    while (cin >> n >> m) {
          int cnt = 0;
        for (int i = 0; i < m; i++) {
              int x;
            cin >> x;
            if (x > 0) a[cnt++] = x;
        }
        m = cnt;
        LL res = 0;
        n--;
        for (int i = 0; i < m; i++) {
              dfs(i, 1, 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[12];

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 = 0;
        for (int i = 0; i < m; i++) {
            int x;
            cin >> x;
            if (x > 0 && x <= n) a[d++] = x;
        }

        n--;
        LL res = 0;
        for (LL i = 1; i < (1 << d); i++) {
              int num = 0;
            for (LL j = i; j != 0; j >>= 1) num += j & 1;
            LL lcm = 1;
            for (LL j = 0; j < d; j++) {
                  if ((i >> j) & 1) {
                    lcm = lcm / gcd(lcm, a[j]) * a[j];
                    if (lcm > n) break;
                }
            }
            if (num % 2 == 0) res -= n / lcm;
            else res += n / lcm;
        }
    
        cout << res << endl;
    }
    return 0;
}

 

容斥原理学习(Hdu 4135,Hdu 1796)

标签:des   style   blog   http   color   os   io   java   ar   

原文地址:http://www.cnblogs.com/Stomach-ache/p/3959117.html

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