容斥原理是对集合的运算。
举个例子三个集合的容斥原理： A ∪ B ∪ C = A + B + C - A ∩ B - B ∩ C - C ∩ A + A ∩ B ∩ C.
多个集合的容斥原理：
hdu 1796
How many integers can you find
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.
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
Author
wangye
在总数以下的所有数为一个集合要寻找集合中所求子集。
子集是所有给出数字的倍数集合的并集，典型的容斥原理。
用DFS的思想来举出所有集合的情况，有奇数个数的集合中的元素数加进去，有偶数个数的集合中的元素数减价去，最后即可求出并集。
刚开始交代码时候 judge结果是
Runtime Error
(INTEGER_DIVIDE_BY_ZERO)
这个略坑，所给的数竟然有0，改掉以后就变WA了，就知道代码有点挫，有可能超int了，就改成__int64就过了。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 15
#define INF 1<<25
typedef long long ll;
using namespace std;
__int64 num,ans;
__int64 aa[maxn],lim,tot,s;
bool vv[maxn];
__int64 gcd(__int64 x,__int64 y)
{
return !y?x:gcd(y,x%y);
}
__int64 lcm(__int64 x,__int64 y)
{
return x*y/gcd(x,y);
}
__int64 dfs(__int64 x,__int64 y)
{
vv[x]=1;
s++;
if(s==y)
{
__int64 rec=1;
for(int i=0; i<tot; i++)
if(vv[i])
rec=lcm(rec,aa[i]);
if(y%2)
ans+=lim/rec;
else ans-=lim/rec;
return 0;
}
for(int i=x; i<tot; i++)
{
if(!vv[i])
{
dfs(i,y);
vv[i]=0;
s--;
}
}
}
int main()
{
while(scanf("%I64d%I64d",&lim,&tot)!=EOF)
{
memset(aa,0,sizeof(aa));
lim--;
int ii=0;
for(int i=0; i<tot; i++)
{
int xx;
scanf("%d",&xx);
if(xx)
aa[ii++]=xx;
}
tot=ii;
ans=0;
for(int i=1; i<=tot; i++)
{
for(int j=0; j<tot; j++)
{
memset(vv,0,sizeof(vv));
s=0;
dfs(j,i);
}
}
printf("%I64d\n",ans);
}
}