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 }