多重集的組合數公式得記下。cf451E就是這個的裸題

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;

int n;LL m;

LL quick_pow(LL A,LL p)
{
    LL ret
 
=1;
    while(p!=0)
    {
        if(p%2==1)ret=(ret*A)%mod;
        A=(A*A)%mod;p/=2;
    }
    return ret;
}
LL jiecheng(LL k)
{
    LL ret=1;
    for(int i=1;i<=k;i++)ret=(ret*i)%mod;
    return ret;
}

LL a[30];
LL calc(int zt)
{
    LL u=n+m-1,cnt=0;
    for(int i=0;i<n;i++)
        
 
if(zt&(1<<i)) u-=a[i], cnt++;
    u-=cnt;
    if(u<n-1)return 0;
    
    LL ret=1;
    for(int i=1;i<=n-1;i++)
    {
        ret=(ret*(u%mod))%mod, u--;
    }
    return (cnt%2==0)?ret:-ret;
}

int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    scanf(
 
"%d%lld",&n,&m);
    for(int i=0;i<n;i++)scanf("%lld",&a[i]);
    
    int li=(1<<n)-1; LL ans=0,ny=quick_pow(jiecheng(n-1),mod-2);
    for(int zt=0;zt<=li;zt++)
    {
        ans=(ans+calc(zt)*ny%mod)%mod;
    }
    printf("%lld\n",(ans+mod)%mod);
    return 0;
}

bzoj1101 沒什麽難度。

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

bool v[51000];
int pr,prime[51000],u[51000];
void mobius_inversion()
{
    u[1]=1;pr=0;
    memset(v,true,sizeof(v));
    for(int i=2;i<=50010;i++)
    {
        if(v[i]==true)
        {
            prime[++pr]=i;
            u[i]=-1;
        }
        for(int j=1;j<=pr&&i*prime[j]<=50010;j++)
        {
            v[i*prime[j]]=false;
            if(i%prime[j]==0)
            {
                u[i*prime[j]]=0;
                break;
            }
            else u[i*prime[j]]=-u[i];
        }
        u[i]+=u[i-1];
    }
}
int solve(int n,int m)
{
    int li=min(n,m),last,ans=0;
    for(int i=1;i<=li;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        ans+=(u[last]-u[i-1])*(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==0){printf("0\n");continue;}
        a/=d;b/=d;
        printf("%d\n",solve(a,b));
    }
    return 0;
}

0x37 容斥原理與莫比烏斯函數