[51NOD]2 3 5 7的倍数（容斥原理）

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1284

似乎最近没怎么做容斥原理啊，来复习一发概念。

容斥原理公式：

一般做法都是先搞一个式子出来，再枚举所有子集，用式子来确定符号和子集的关系。小学生容斥。。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%I64d", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onecnt(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70 
 71 LL ret;
 72 LL n;
 73 
 74 int main() {
 75     // FRead();
 76     while(cin >> n) {
 77         LL a,b,c,d,ab,ac,ad,bc,bd,cd,abc,abd,acd,bcd,abcd;
 78         a = n / 2;
 79         b = n / 3;
 80         c = n / 5;
 81         d = n / 7;
 82 
 83         ab = n / 6;
 84         ac = n / 10;
 85         ad = n / 14;
 86         bc = n / 15;
 87         bd = n / 21;
 88         cd = n / 35;
 89 
 90         abc = n / 30;
 91         abd = n / 42;
 92         acd = n / 70;
 93         bcd = n / 105;
 94 
 95         abcd = n / 210;
 96 
 97         cout << n-(a+b+c+d-ab-ac-ad-bc-bd-cd+abc+abd+acd+bcd-abcd) << endl;
 98     }
 99     RT 0;
100 }

 

posted @ 2016-09-03 16:44  Kirai  阅读(234)  评论(0)  编辑  收藏  举报