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  • 【BZOJ 3157&&BZOJ 3516】—国王奇遇记&&加强版(扰动法+Dp)

    传送门

    加强版

    考虑设f(i)=j=1njimjf(i)=sum_{j=1}^nj^im^j
    用扰动法化简,得到
    (m1)f(i)=j=1njimj+1j=1njimj=j=1n+1(j1)imjj=1njimj=nimn+1+j=1nmjk=0i1(ik)(1)ikjk=nimn+1+k=0i1(ik)(1)ikj=1njkmj=nimn+1+k=0i1(ik)(1)ikf(k)(m - 1) * f(i) = sum_{j=1}^n j^i * m^{j + 1} - sum_{j=1}^n j^i * m^j \ = sum_{j=1}^{n + 1} (j - 1)^i * m^j - sum_{j=1}^n j^i * m^j \ = n^i * m^{n + 1} + sum_{j=1}^n m^j sum_{k = 0}^{i - 1} {i choose k} * (-1)^{i - k} * j^k \ = n^i * m^{n + 1} + sum_{k = 0}^{i - 1} {i choose k} * (-1)^{i - k} sum_{j = 1}^n j^k * m^j \ = n^i * m^{n + 1} + sum_{k = 0}^{i - 1} {i choose k} * (-1)^{i - k} * f(k)

    然后O(m2)dpO(m^2)dp就完了

    注意特判m=1m=1的情况

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<20|1;
    inline char gc(){
    	static char ibuf[RLEN],*ib,*ob;
    	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    	return (ib==ob)?EOF:*ib++;
    }
    #define gc getchar
    inline int read(){
    	char ch=gc();
    	int res=0,f=1;
    	while(!isdigit(ch))f^=ch=='-',ch=gc();
    	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    	return f?res:-res;
    }
    #define re register
    #define pb push_back
    #define cs const
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define poly vector<int>
    #define bg begin
    cs int mod=1e9+7;
    inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
    inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
    inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
    inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
    inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
    inline void Mul(int &a,int b){a=mul(a,b);}
    inline int ksm(int a,int b,int res=1){
    	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
    }
    inline void chemx(int &a,int b){a<b?a=b:0;}
    inline void chemn(int &a,int b){a>b?a=b:0;}
    cs int N=1006;
    int fac[N],ifac[N];
    inline void init(cs int len=N-5){
    	fac[0]=ifac[0]=1;
    	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
    	ifac[len]=ksm(fac[len],mod-2);
    	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
    }
    inline int C(int n,int m){
    	if(n<m)return 0;
    	return mul(fac[n],mul(ifac[m],ifac[n-m]));
    }
    int n,m;
    int f[N],pt[N];
    int main(){
    	init(),n=read(),m=read();
    	if(m==1){cout<<(1ll*(n+1)*n/2)%mod;return 0;}
    	pt[0]=1;int mt=ksm(m,n+1),inv=ksm(m-1,mod-2);
    	for(int i=1;i<N;i++)pt[i]=mul(pt[i-1],n);
    	f[0]=mul(dec(mt,m),inv);
    	for(int i=1;i<=m;i++){
    		int res=0;
    		for(int k=0;k<=i-1;k++)
    			if((i-k)&1)Dec(res,mul(f[k],C(i,k)));
    			else Add(res,mul(f[k],C(i,k)));
    		Add(res,mul(pt[i],mt));
    		f[i]=mul(res,inv);
    	}
    	cout<<f[m];
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328645.html
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