HDU 1796 How many integers can you find 容斥原理
How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4848 Accepted Submission(s): 1388
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
Output For each case, output the number.
Sample Input
12 2
2 3
Sample Output
7
Author wangye
Source 2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）
已知n和m，有m個元素，求小於n且是m個元素中任意元素的倍數的個數。 基礎容斥:ans=整除1個元素個數-整除2個元素個數+整除3個元素個數-整除4個元素個數+....
//904MS 1596K
#include
#include
int s[27],vis[27],sum,n,m,k;
int gcd(int a,int b)//最大公約數
{
return b?gcd(b,a%b):a;
}
int lcm(int a,int b)//最小公倍數
{
return a/gcd(a,b)*b;
}
void dfs(int x,int ans,int now)//x代表當前第幾個數，ans代表一共ans個數求lcm,now代表當前有now個數
{
if(now==ans)
{
int a=1;
for(int i=1;i<=k;i++)
if(vis[i])a=lcm(a,s[i]);
if(ans&1)sum+=(n-1)/a;
else sum-=(n-1)/a;
return ;
}
for(;x<=k;x++)
if(!vis[x])
{
vis[x]=1;
dfs(x+1,ans,now+1);
vis[x]=0;
}
}
int main()
{
while(scanf(%d%d,&n,&m)!=EOF)
{
int a;
k=0,sum=0;
for(int i=1;i<=m;i++)
{
scanf(%d,&a);
if(a)s[++k]=a;
}
for(int i=1;i<=k;i++)
{
memset(vis,0,sizeof(vis));
dfs(1,i,0);
}
printf(%d
,sum);
}
return 0;
}
