HDU 1796 How many integers can you find 容斥原理
2016-11-17 20:16
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题意:
给定一个数字n,求“小于n”的所有数字中,有多少个满足可以整数给出的m个元素
思路:
很明显是个高中的容斥原理,自己写了个dfs,结果忘记剔除0,1,以及比如4,8有公因子的特殊条件了。心酸。代码如下
#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll sum;
ll n,m,num;
int cnt=0;
ll a[105];
int vis[105];
ll gcd(ll x,ll y)
{
return y==0?x:gcd(y,x%y);
}
void dfs(int cur,long long lcm,int ced)
{
lcm=a[cur]/gcd(a[cur],lcm)*lcm;
if(ced&1)
sum+=n/lcm;
else
sum-=n/lcm;
for(int i=cur+1; i<=cnt; i++)
dfs(i,lcm,ced+1);
}
int main()
{
while(cin>>n>>m)
{
n--;
cnt=0;
int x;
while(m--)
{
scanf("%d",&x);
if(x!=0)
a[++cnt]=x;
}
sum=0;
for(int i=1;i<=cnt;i++)
dfs(i,1,1);
cout<<sum<<endl;
}
return 0;
}
给定一个数字n,求“小于n”的所有数字中,有多少个满足可以整数给出的m个元素
思路:
很明显是个高中的容斥原理,自己写了个dfs,结果忘记剔除0,1,以及比如4,8有公因子的特殊条件了。心酸。代码如下
#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll sum;
ll n,m,num;
int cnt=0;
ll a[105];
int vis[105];
ll gcd(ll x,ll y)
{
return y==0?x:gcd(y,x%y);
}
void dfs(int cur,long long lcm,int ced)
{
lcm=a[cur]/gcd(a[cur],lcm)*lcm;
if(ced&1)
sum+=n/lcm;
else
sum-=n/lcm;
for(int i=cur+1; i<=cnt; i++)
dfs(i,lcm,ced+1);
}
int main()
{
while(cin>>n>>m)
{
n--;
cnt=0;
int x;
while(m--)
{
scanf("%d",&x);
if(x!=0)
a[++cnt]=x;
}
sum=0;
for(int i=1;i<=cnt;i++)
dfs(i,1,1);
cout<<sum<<endl;
}
return 0;
}
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