0x37 容斥原理与莫比乌斯函数

多重集的组合数公式得记下。cf451E就是这个的裸题

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL mod=1e9+;

int n;LL m;

LL quick_pow(LL A,LL p)
{
    LL ret=;
    while(p!=)
    {
        if(p%==)ret=(ret*A)%mod;
        A=(A*A)%mod;p/=;
    }
    return ret;
}
LL jiecheng(LL k)
{
    LL ret=;
    for(int i=;i<=k;i++)ret=(ret*i)%mod;
    return ret;
}

LL a[];
LL calc(int zt)
{
    LL u=n+m-,cnt=;
    for(int i=;i<n;i++)
        if(zt&(<<i)) u-=a[i], cnt++;
    u-=cnt;
    if(u<n-)return ;

    LL ret=;
    for(int i=;i<=n-;i++)
    {
        ret=(ret*(u%mod))%mod, u--;
    }
    return (cnt%==)?ret:-ret;
}

int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    scanf("%d%lld",&n,&m);
    for(int i=;i<n;i++)scanf("%lld",&a[i]);

    int li=(<<n)-; LL ans=,ny=quick_pow(jiecheng(n-),mod-);
    for(int zt=;zt<=li;zt++)
    {
        ans=(ans+calc(zt)*ny%mod)%mod;
    }
    printf("%lld\n",(ans+mod)%mod);
    return ;
}

cf451E

bzoj1101 没什么难度。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

bool v[];
int pr,prime[],u[];
void mobius_inversion()
{
    u[]=;pr=;
    memset(v,true,sizeof(v));
    for(int i=;i<=;i++)
    {
        if(v[i]==true)
        {
            prime[++pr]=i;
            u[i]=-;
        }
        for(int j=;j<=pr&&i*prime[j]<=;j++)
        {
            v[i*prime[j]]=false;
            if(i%prime[j]==)
            {
                u[i*prime[j]]=;
                break;
            }
            else u[i*prime[j]]=-u[i];
        }
        u[i]+=u[i-];
    }
}
int solve(int n,int m)
{
    int li=min(n,m),last,ans=;
    for(int i=;i<=li;i=last+)
    {
        last=min(n/(n/i),m/(m/i));
        ans+=(u[last]-u[i-])*(n/i)*(m/i);
    }
    return ans;
}
int main()
{
    mobius_inversion();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int a,b,d;
        scanf("%d%d%d",&a,&b,&d);
        if(d==){printf("0\n");continue;}
        a/=d;b/=d;
        printf("%d\n",solve(a,b));
    }
    return ;
}

bzoj1101