hdu 6036 NTT取模（板子）+组合数学

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HDU 6036 - Division Game [ 组合数学,NTT ]  |  2017 Multi-University Training Contest 1

题意：

    k堆石子围成一个圈，数量均为n，编号为0至k-1

    第i轮可以操作第 (i+1) mod k 堆石子，必须拿石子且原石子数量要求整除操作后石子数量

    任意一堆石子只剩一颗后停止游戏，问游戏停止在第i堆的方案数

    限制:

        n很大，按唯一分解定理形式给出 n = p1^e1 * p2^e2 * ... * pm^em

        m,k <= 10    ∑ei <= 1e5

分析：

    设 w = ∑ei ,则每堆石子最多操作w次

    设 F(x) 为一个堆操作 x 次恰好变为1的方案数，则一个堆操作 x-1 次后不变为 1 的方案数也为 F(x)

    对于石子堆i的方案数，枚举总操作次数x，设此时方案数为 ans[i][x]

    则显然当石子堆i操作x次时，0 < j < i 的石子堆 j 均操作x次，而i < j <= k 的石子堆均操作x-1

    故 ans[i][x] = [0<j<i的堆操作x次后不为1] * [堆i操作x次恰好为1] * [i<j<=k的堆操作x-1次后不为1]

                 = F(x+1)^(i-1) * F(x) * F(x)^(k-i)

                 = F(x+1)^(i-1) * F(x)^(k-i+1)

     

    现在研究一下上式 x 的取值范围

        由于每个堆至多操作 w 次，则 [0<j<i的堆操作x次后不为1] 的 x ∈[0 , w-1]

                                    [堆i操作x次恰好为1] 的 x ∈[0 , w]

                                    [i<j<=k的堆操作x-1次后不为1] 的 x ∈[0 , w+1]

        易得出结论 x ∈[0 , w-1]

        但当 i = 1 时，0<j<i的堆 不存在，故其可以取到 x = w

        所以分类讨论：

            当 i = 1 时，ans[i][x] = F(x+1)^(i-1) * F(x)^(k-i+1) + F(w) , x ∈[0 , w-1]

            当 i > 1 时，ans[i][x] = F(x+1)^(i-1) * F(x)^(k-i+1) , x ∈[0 , w-1]

 

    研究如何求出 F(x)：

        此时有两个限制条件

            1. 对于每个质因子pi，在 x 次内取完

                即ei个相同的数字分成x组，允许空组，根据挡板法，方案数为 Comb(ei+x-1, x-1)

                根据乘法原理 总方案数 F'(x) = ∏ Comb(ei+x-1, x-1) [1<=i<=m]

                但由于存在第二个限制，F(x) != F'(x)

            2. 每一次至少存在一个质因子被取

                则根据容斥原理

                F(x) = 随意取的方案数 - 某一次什么都没有取的方案数

                                      + 某两次什么都没有取的方案数

                                      - 某三次什么都没有取的方案数

                                      + ...

                由于 x次中k次没有取的方案数 = Comb(x,k) * x-k次恰好取完的方案数 = Comb(x,k) *F(x-k)

                则 F(x) = F'(x) - Comb(x,1) * F(x-1)

                                + Comb(x,2) * F(x-2)

                                - Comb(x,3) * F(x-3)

                                + ...

                        = Σ (-1)^i * C(x, i) * F(x-i) [0<=i<x]

                将组合数打开优化：

                    F(x) = Σ (-1)^i * x! / i! / (x-i)! * F(x-i) [0<=i<x]

                    F(x)/x!  =  Σ (-1)^i/i!  *  F(x-i)/(x-i)!  [0<=i<x]

                可以看出是卷积，再考虑模数特殊，用 NTT 优化

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e3+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;const int N = 100005;const int MOD =  985661441;int e[20],m,k,n;int g[N],ans[20];int pa[2][N];namespace NTT{const int G = 3;const int NUM = 20;int wn[20];int mul(int x, int y){return (LL)x*y%MOD;}int PowMod(int a, int b){int res = 1;a %= MOD;while (b){if (b&1) res = mul(res, a);a = mul(a, a);b >>= 1;}return res;}void GetWn(){for (int i = 0; i < NUM; i++){int t = 1<<i;wn[i] = PowMod(G, (MOD-1)/t);}}void Change(int a[], int len){int i, j, k;for (i = 1, j = len/2; i < len-1; i++){if (i < j) swap(a[i], a[j]);k = len/2;while (j >= k){j -= k;k /= 2;}if (j < k) j += k;}}void NTT(int a[], int len, int on){Change(a, len);int id = 0;for (int h = 2; h <= len; h <<= 1){id++;for (int j = 0; j < len; j += h){int w = 1;for (int k = j; k < j + h/2; k++){int u = a[k] % MOD;int t = mul(a[k+h/2], w);a[k] = (u+t) % MOD;a[k+h/2] = ((u-t)%MOD + MOD) % MOD;w = mul(w, wn[id]);}}}if (on == -1){for (int i = 1; i < len/2; i++)swap(a[i], a[len-i]);int inv = PowMod(len, MOD-2);for (int i = 0; i < len; i++)a[i] = mul(a[i], inv);}}}int a[N<<3],b[N<<3];namespace COMB{int F[N<<1], Finv[N<<1], inv[N<<1];void init(){inv[1] = 1;for (int i = 2; i < N<<1; i++){inv[i] = (LL)(MOD - MOD/i) * inv[MOD%i] % MOD;}F[0] = Finv[0] = 1;for (int i = 1; i < N<<1; i++){F[i] = (LL)F[i-1] * i % MOD;Finv[i] = (LL)Finv[i-1] * inv[i] % MOD;}}int comb(int n, int m){if (m < 0 || m > n) return 0;return (LL)F[n] * Finv[n-m] % MOD * Finv[m] % MOD;}}using namespace COMB;int main(){    int cas=1,len;    NTT::GetWn();    init();    W(s_2(m,k)!=EOF)    {        n=0;        FOR(1,m,i)        {            scanf("%*d%d",&e[i]);            n+=e[i];        }        ++n;//sigma e_i        FOr(0,n,x)        {            g[x]=1;            FOR(1,m,i)                g[x]=(LL)g[x]*comb(e[i]+x-1,x-1)%MOD;        }        FOr(0,n,i)        {            a[i]=i%2?(MOD-Finv[i]):Finv[i];//  (-1)^i/i!            b[i]=(LL)g[i]*Finv[i]%MOD;//F(x-i)/(x-i)!        }        len = 1;        while (len < n*2) len <<= 1;//get_len        for(int i=n;i<len;i++) a[i]=b[i]=0;        NTT::NTT(a,len,1);        NTT::NTT(b,len,1);        for(int i=0;i<len;i++) a[i]=NTT::mul(a[i],b[i]);        NTT::NTT(a,len,-1);        for(int i=0;i<n;i++)            a[i]=(LL)a[i]*F[i]%MOD;        me(ans,0);        a[n]=0;        int pre=1,cur=0;        FOr(1,n,x)        {            pre^=1,cur^=1;//优化，减少内存使用            pa[cur][0]=1;            FOR(1,k,i)                pa[cur][i]=(LL)pa[cur][i-1]*a[x]%MOD;            if(x>1)                FOR(1,k,i)                    ans[i]=(ans[i]+(LL)pa[cur][i-1]*pa[pre][k-i+1])%MOD;        }        ans[1]=(ans[1]+pa[cur][k])%MOD;        printf("Case #%d:",cas++);        FOR(1,k,i) printf(" %d",ans[i]);        puts("");    }}