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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): 531Problem DescriptionGiven 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.
InputThe 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).
OutputFor 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 Input21 10 23 15 5
Sample OutputCase #1: 5Case #2: 10HintIn 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): 1198Problem DescriptionNow 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.
InputThere 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.
OutputFor each case, output the number.
Sample Input12 22 3
Sample Output7
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）
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