Forcing the composition formula sequence-- + Significance

sequence

Regardless of the length <Program Number F = x a (x), and (F (x) -F (x-1)) * x contribution to the answer in

n squares approach can be directly DP,

 

Feel equation at all,

Launch formula: Similarly coat, each of length i of the i th calculation.

Then the inclusion-exclusion under:

F is the number of the program, he asked:

 Number of segments divided enumeration, the number i exceeds enumeration for inclusion and exclusion:

Breakthrough: There is a ni * k-1, means that i * k <= n, so i k violence and enumerate total nlogn complexity !

Proposed, considering kill j

Forcing the formula:

deal with:

(How to see how I can not see what the truth looks like)

Meaning it came to a combination of:

There are n-ik balls, we start with the j-th selected, and then select from the k selected in the j. In j balls, we selected a special ball for the rest of the balls m 1-color staining.

Consider discussing this particular ball is not a ball in the k

Can be obtained;

(Here less write C (ni * kk, k))

Power pretreatment m-1 and m-th power of the inverse element factorial factorial

O (nlogn)

Remember Finally n * m ^ n-

#include<bits/stdc++.h>
#define reg register int
#define il inline
#define fi first
#define se second
#define mk(a,b) make_pair(a,b)
#define numb (ch^'0')
#define pb push_back
#define solid const auto &
#define enter cout<<endl
#define pii pair<int,int>
using namespace std;
typedef long long ll;
template<class T>il void rd(T &x){
    char ch;x=0;bool fl=false;while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
    for(x=numb;isdigit(ch=getchar());x=x*10+numb);(fl==true)&&(x=-x);}
template<class T>il void output(T x){if(x/10)output(x/10);putchar(x%10+'0');}
template<class T>il void ot(T x){if(x<0) putchar('-'),x=-x;output(x);putchar(' ');}
template<class T>il void prt(T a[],int st,int nd){for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');}

int mod;
namespace Modulo{
int ad(int x,int y){return (x+y)>=mod?x+y-mod:x+y;}
void inc(int &x,int y){x=ad(x,y);}
int mul(int x,int y){return (ll)x*y%mod;}
void inc2(int &x,int y){x=mul(x,y);}
int qm(int x,int y=mod-2){int ret=1;while(y){if(y&1) ret=mul(x,ret);x=mul(x,x);y>>=1;}return ret;}
}
using namespace Modulo;
namespace Miracle{
const int N=300000+5;
int n,m;
int jie[N],inv[N];
int iv[N];
int m0[N],m1[N];
int C(int n,int m){
    if(n<0||m<0||n<m) return 0;
    return mul(jie[n],mul(inv[m],inv[n-m]));
}
int main(){
    rd(n);rd(m);rd(mod);
    jie[0]=1;
    for(reg i=1;i<=n;++i) jie[i]=mul(jie[i-1],i);
    inv[n]=qm(jie[n],mod-2);
    for(reg i=n-1;i>=0;--i) inv[i]=mul(inv[i+1],i+1);
    iv[1]=1;
    for(reg i=2;i<=n;++i){
        iv[i]=mul(mod-mod/i,iv[mod%i]);
    }
    m0[0]=m1[0]=1;
    for(reg i=1;i<=n;++i){
        m0[i]=mul(m0[i-1],m);
        m1[i]=mul(m1[i-1],m-1);
    }
    int ans=0;
    for(reg i=0;i<n;++i){
        for(reg k=0;k<=n;++k){
            if(i*k+k>n) break;
            int tmp=0;//
            if(k!=0) tmp=ad(tmp,mul(C(n-i*k,k),mul(k,mul(m1[k-1],m0[n-i*k-k]))));
            if(n-i*k-k-1>=0) tmp=ad(tmp,mul(C(n-i*k,k),mul(m1[k],mul(m0[n-i*k-k-1],n-i*k-k))));
            tmp=mul(tmp,(k&1)?mod-1:1);
            tmp=mul(tmp,iv[n-i*k]);
            ans=ad(ans,tmp);
        }
    }
    ans=mul(ans,m);
    ans=ad(mul(n,qm(m,n)),mod-ans);
    ot(ans);
    return 0;
}

}
signed main(){
    Miracle::main();
    return 0;
}

/*
   Author: *Miracle*
*/

 Is a breakthrough i, k of the total number of nlogn level, kill j with a combination of the strong sense of derivation