Combinatorial Number
一、组合数取模1：
例题：回答T组询问，输出\(C_{n}^{m} \bmod 10^9+7\)的值。
\(C_{n}^{m} = \dfrac{n!}{m!*(n-m)!}\)
\(\dfrac{a}{b} \bmod p\) ==> \(a \bmod p*(b \bmod p)^{-1}\)
即
-
\(n!*(m!)^{-1}*(n-m)^{-1}\)
-
\(inv[i] = (p-\dfrac{p}{i})*inv[p\)%\(i]\)%\(p\)
代码
#include<bits/stdc++.h> using namespace std; const int N = 1e7+10; const int mod = 1e9+7; typedef long long ll; ll fac[N],fnv[N]; ll ksm(ll a,ll b) { ll ans = 1,base = a; while(b>0) { if(b&1) { ans *= base; ans %= mod; } base *= base; base%=mod; b>>=1; } return ans; } ll binom(int n,int m) { if(m<0||m>n)return 0; return fac[n]*fnv[m]%mod*fnv[n-m]%mod; } int main() { fac[0] = 1; for(int i = 1;i<=N;i++) fac[i] = fac[i-1]*i%mod; fnv[N] = ksm(fac[N],mod-2); for(int i = N;i>=1;i--) fnv[i-1] = fnv[i]*i%mod; assert(fnv[0]==1); int t; cin>>t; while(t--) { int a,b; cin>>a>>b; cout<<binom(a,b)<<endl; } return 0; }
二、组合数取模2：分段打表求组合数
例题：回答\(T\)组询问，输出\(C_{n}^{m} \bmod10^9+7\)的值。
对于所有数据，保证\(1≤T≤10\),\(1≤m≤n≤10^9\)
思路：
分段打表,每隔\(10^6\)的距离就打表
\(0!,10^6! ,2*10^6!,...,n!\)
比如我要求\((10^8+12345)!\)
可以先查表找到\(10^8\)的阶乘,再求\(12345\)次阶乘,这样\(for\)一次最多\(10^6\)
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int T = 1e6;
const int mod = 1e9+7;
ll fact[] = {1,641102369,578095319,5832229,259081142,974067448,316220877,690120224,251368199,980250487,682498929,134623568,95936601,933097914,167332441,598816162,336060741,248744620,626497524,288843364,491101308,245341950,565768255,246899319,968999,586350670,638587686,881746146,19426633,850500036,76479948,268124147,842267748,886294336,485348706,463847391,544075857,898187927,798967520,82926604,723816384,156530778,721996174,299085602,323604647,172827403,398699886,530389102,294587621,813805606,67347853,497478507,196447201,722054885,228338256,407719831,762479457,746536789,811667359,778773518,27368307,438371670,59469516,5974669,766196482,606322308,86609485,889750731,340941507,371263376,625544428,788878910,808412394,996952918,585237443,1669644,361786913,480748381,595143852,837229828,199888908,526807168,579691190,145404005,459188207,534491822,439729802,840398449,899297830,235861787,888050723,656116726,736550105,440902696,85990869,884343068,56305184,973478770,168891766,804805577,927880474,876297919,934814019,676405347,567277637,112249297,44930135,39417871,47401357,108819476,281863274,60168088,692636218,432775082,14235602,770511792,400295761,697066277,421835306,220108638,661224977,261799937,168203998,802214249,544064410,935080803,583967898,211768084,751231582,972424306,623534362,335160196,243276029,554749550,60050552,797848181,395891998,172428290,159554990,887420150,970055531,250388809,487998999,856259313,82104855,232253360,513365505,244109365,1559745,695345956,261384175,849009131,323214113,747664143,444090941,659224434,80729842,570033864,664989237,827348878,195888993,576798521,457882808,731551699,212938473,509096183,827544702,678320208,677711203,289752035,66404266,555972231,195290384,97136305,349551356,785113347,83489485,66247239,52167191,307390891,547665832,143066173,350016754,917404120,296269301,996122673,23015220,602139210,748566338,187348575,109838563,574053420,105574531,304173654,542432219,34538816,325636655,437843114,630621321,26853683,933245637,616368450,238971581,511371690,557301633,911398531,848952161,958992544,925152039,914456118,724691727,636817583,238087006,946237212,910291942,114985663,492237273,450387329,834860913,763017204,368925948,475812562,740594930,45060610,806047532,464456846,172115341,75307702,116261993,562519302,268838846,173784895,243624360,61570384,481661251,938269070,95182730,91068149,115435332,495022305,136026497,506496856,710729672,113570024,366384665,564758715,270239666,277118392,79874094,702807165,112390913,730341625,103056890,677948390,339464594,167240465,108312174,839079953,479334442,271788964,135498044,277717575,591048681,811637561,353339603,889410460,839849206,192345193,736265527,316439118,217544623,788132977,618898635,183011467,380858207,996097969,898554793,335353644,54062950,611251733,419363534,965429853,160398980,151319402,990918946,607730875,450718279,173539388,648991369,970937898,500780548,780122909,39052406,276894233,460373282,651081062,461415770,358700839,643638805,560006119,668123525,686692315,673464765,957633609,199866123,563432246,841799766,385330357,504962686,954061253,128487469,685707545,299172297,717975101,577786541,318951960,773206631,306832604,204355779,573592106,30977140,450398100,363172638,258379324,472935553,93940075,587220627,776264326,793270300,291733496,522049725,579995261,335416359,142946099,472012302,559947225,332139472,499377092,464599136,164752359,309058615,86117128,580204973,563781682,954840109,624577416,895609896,888287558,836813268,926036911,386027524,184419613,724205533,403351886,715247054,716986954,830567832,383388563,68409439,6734065,189239124,68322490,943653305,405755338,811056092,179518046,825132993,343807435,985084650,868553027,148528617,160684257,882148737,591915968,701445829,529726489,302177126,974886682,241107368,798830099,940567523,11633075,325334066,346091869,115312728,473718967,218129285,878471898,180002392,699739374,917084264,856859395,435327356,808651347,421623838,105419548,59883031,322487421,79716267,715317963,429277690,398078032,316486674,384843585,940338439,937409008,940524812,947549662,833550543,593524514,996164327,987314628,697611981,636177449,274192146,418537348,925347821,952831975,893732627,1277567,358655417,141866945,581830879,987597705,347046911,775305697,125354499,951540811,247662371,343043237,568392357,997474832,209244402,380480118,149586983,392838702,309134554,990779998,263053337,325362513,780072518,551028176,990826116,989944961,155569943,596737944,711553356,268844715,451373308,379404150,462639908,961812918,654611901,382776490,41815820,843321396,675258797,845583555,934281721,741114145,275105629,666247477,325912072,526131620,252551589,432030917,554917439,818036959,754363835,795190182,909210595,278704903,719566487,628514947,424989675,321685608,50590510,832069712,198768464,702004730,99199382,707469729,747407118,302020341,497196934,5003231,726997875,382617671,296229203,183888367,703397904,552133875,732868367,350095207,26031303,863250534,216665960,561745549,352946234,784139777,733333339,503105966,459878625,803187381,16634739,180898306,68718097,985594252,404206040,749724532,97830135,611751357,31131935,662741752,864326453,864869025,167831173,559214642,718498895,91352335,608823837,473379392,385388084,152267158,681756977,46819124,313132653,56547945,442795120,796616594,256141983,152028387,636578562,385377759,553033642,491415383,919273670,996049638,326686486,160150665,141827977,540818053,693305776,593938674,186576440,688809790,565456578,749296077,519397500,551096742,696628828,775025061,370732451,164246193,915265013,457469634,923043932,912368644,777901604,464118005,637939935,956856710,490676632,453019482,462528877,502297454,798895521,100498586,699767918,849974789,811575797,438952959,606870929,907720182,179111720,48053248,508038818,811944661,752550134,401382061,848924691,764368449,34629406,529840945,435904287,26011548,208184231,446477394,206330671,366033520,131772368,185646898,648711554,472759660,523696723,271198437,25058942,859369491,817928963,330711333,724464507,437605233,701453022,626663115,281230685,510650790,596949867,295726547,303076380,465070856,272814771,538771609,48824684,951279549,939889684,564188856,48527183,201307702,484458461,861754542,326159309,181594759,668422905,286273596,965656187,44135644,359960756,936229527,407934361,267193060,456152084,459116722,124804049,262322489,920251227,816929577,483924582,151834896,167087470,490222511,903466878,361583925,368114731,339383292,388728584,218107212,249153339,909458706,322908524,202649964,92255682,573074791,15570863,94331513,744158074,196345098,334326205,9416035,98349682,882121662,769795511,231988936,888146074,137603545,582627184,407518072,919419361,909433461,986708498,310317874,373745190,263645931,256853930,876379959,702823274,147050765,308186532,175504139,180350107,797736554,606241871,384547635,273712630,586444655,682189174,666493603,946867127,819114541,502371023,261970285,825871994,126925175,701506133,314738056,341779962,561011609,815463367,46765164,49187570,188054995,957939114,64814326,933376898,329837066,338121343,765215899,869630152,978119194,632627667,975266085,435887178,282092463,129621197,758245605,827722926,201339230,918513230,322096036,547838438,985546115,852304035,593090119,689189630,555842733,567033437,469928208,212842957,117842065,404149413,155133422,663307737,208761293,206282795,717946122,488906585,414236650,280700600,962670136,534279149,214569244,375297772,811053196,922377372,289594327,219932130,211487466,701050258,398782410,863002719,27236531,217598709,375472836,810551911,178598958,247844667,676526196,812283640,863066876,857241854,113917835,624148346,726089763,564827277,826300950,478982047,439411911,454039189,633292726,48562889,802100365,671734977,945204804,508831870,398781902,897162044,644050694,892168027,828883117,277714559,713448377,624500515,590098114,808691930,514359662,895205045,715264908,628829100,484492064,919717789,513196123,748510389,403652653,574455974,77123823,172096141,819801784,581418893,15655126,15391652,875641535,203191898,264582598,880691101,907800444,986598821,340030191,264688936,369832433,785804644,842065079,423951674,663560047,696623384,496709826,161960209,331910086,541120825,951524114,841656666,162683802,629786193,190395535,269571439,832671304,76770272,341080135,421943723,494210290,751040886,317076664,672850561,72482816,493689107,135625240,100228913,684748812,639655136,906233141,929893103,277813439,814362881,562608724,406024012,885537778,10065330,60625018,983737173,60517502,551060742,804930491,823845496,727416538,946421040,678171399,842203531,175638827,894247956,538609927,885362182,946464959,116667533,749816133,241427979,871117927,281804989,163928347,563796647,640266394,774625892,59342705,256473217,674115061,918860977,322633051,753513874,393556719,304644842,767372800,161362528,754787150,627655552,677395736,799289297,846650652,816701166,687265514,787113234,358757251,701220427,607715125,245795606,600624983,10475577,728620948,759404319,36292292,491466901,22556579,114495791,647630109,586445753,482254337,718623833,763514207,66547751,953634340,351472920,308474522,494166907,634359666,172114298,865440961,364380585,921648059,965683742,260466949,117483873,962540888,237120480,620531822,193781724,213092254,107141741,602742426,793307102,756154604,236455213,362928234,14162538,753042874,778983779,25977209,49389215,698308420,859637374,49031023,713258160,737331920,923333660,804861409,83868974,682873215,217298111,883278906,176966527,954913,105359006,390019735,10430738,706334445,315103615,567473423,708233401,48160594,946149627,346966053,281329488,462880311,31503476,185438078,965785236,992656683,916291845,881482632,899946391,321900901,512634493,303338827,121000338,967284733,492741665,152233223,165393390,680128316,917041303,532702135,741626808,496442755,536841269,131384366,377329025,301196854,859917803,676511002,373451745,847645126,823495900,576368335,73146164,954958912,847549272,241289571,646654592,216046746,205951465,3258987,780882948,822439091,598245292,869544707,698611116};
ll fac(int n)
{
ll v = fact[n/T];
for(int i = n/T*T+1;i<=n;i++)
v = v*i%mod;
return v;
}
ll ksm(ll a,ll b)
{
ll ans = 1,base = a;
while(b>0)
{
if(b&1)
{
ans *= base;
ans %= mod;
}
base *= base;
base%=mod;
b>>=1;
}
return ans;
}
ll binom(int n,int m)
{
if(m<0||m>n)return 0;
return fac(n)*ksm(fac(m)*fac(n-m)%mod,mod-2)%mod;
}
int main()
{
//打表
//ll fac = 1;
// for(int i = 1;i< mod;i++)
// {
// fac = fac*i%mod;
// if(i%1000000==0)cout<<fac<<",";
// }
int t;
cin>>t;
while(t--)
{
int a,b;
cin>>a>>b;
cout<<binom(a,b)<<endl;
}
return 0;
}
三、组合数取模3：Lucas
Th(Lucas):
\(0<=ni<=mi<p\)
\(n = n0p^0 + n1p^1 + ... + nkp^k\)
$ m = m0p^0 + m1p^1 + ... + mkp^k$
$ C_{n}^{m} ≡ C_{n0}^{m0} * C_{n1}^{m1} ... C_{nk}^{mk} (\bmod p) $
对于\(C_{a}^{b}\),若\(a<b\) 认为是\(0\)
\[\left( \begin{matrix} 13\\ 4 \end{matrix} \right) \tag{2} \bmod 3
\]
写成三进制
$ \left( \begin{matrix} 111\\ 011 \end{matrix} \right) \tag{2} \bmod 3$ = $ \left( \begin{matrix} 1\\ 0 \end{matrix} \right) \tag{2}$$\left( \begin{matrix} 1\\ 1 \end{matrix} \right) \tag{2}$$ \left( \begin{matrix} 1\\ 1 \end{matrix} \right) \tag{2} \bmod 3$
例题：给定一个素数\(p\)。回答\(T\)组询问，输出\(C_{n}^{m} \bmod p\)的值
\(C_{n}^{m} = \dfrac{n!}{m!*(n-m)!}\)
注意：\(1<=m<=n<=1^{18} 2<=p<=10^{6}\)
\(m,n\)很大，\(p\)比较小的情况
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 1000000;
ll fac[M+10],fnv[M+10],p;
ll ksm(ll a,ll b,ll m)
{
ll ans = 1,base = a;
while(b>0)
{
if(b&1)
{
ans *= base;
ans %= m;
}
base *= base;
base%=m;
b>>=1;
}
return ans;
}
ll binom(ll a,ll b)
{
if(b<0||b>a)return 0;
return fac[a]*fnv[b]%p*fnv[a-b]%p;
}
ll lucas(ll n,ll m,ll p)
{
ll ans = 1;
while(n>0||m>0)//做p进制分解
{
ans = ans*binom(n%p,m%p)%p;
n/=p,m/=p;
}
return ans;
}
int main()
{
int t;
cin>>p>>t;
fac[0] = 1;
for(int i = 1;i<=p-1;i++)
fac[i] = fac[i-1]*i%p;
/*
n!/(m!*(n-m)!)
a/b mod p ==> a mod p*(b mod p)^-1
即:
1.n!*(m!)^-1*(n-m)^-1
2.inv[i] = (p-p/i)*inv[p%i]%p
*/
fnv[p-1] = ksm(fac[p-1],p-2,p);
for(int i = p-1;i>=1;i--)
fnv[i-1] = fnv[i]*i%p;
assert(fnv[0]==1);
while(t--)
{
ll n,m;
cin>>n>>m;
/*
(Lucas)
ll ans = 1;
while(n>0||m>0)//做p进制分解
{
ans = ans*binom(n%p,m%p)%p;
n/=p,m/=p;
}
cout<<ans<<endl;
*/
cout<<lucas(n,m,p)<<endl;
}
}
Tips:
一般也不会直接考你求个组合数,那一般喜欢考\(\bmod 2\)
\(\bmod 2\) 什么意思呢？
\(C_{n}^{m}\) 每一项的n>=m,写成二进制形式只可能是$ \left( \begin{matrix} 1\\ 0 \end{matrix} \right) \tag{2}$$ \left( \begin{matrix} 1\\ 1 \end{matrix} \right) \tag{2}$$ \left( \begin{matrix} 0\\ 0 \end{matrix} \right) \tag{2}$
所以 \(C_{n}^{m} ≡ 1 (\bmod 2)\) 意味着n&m == m
也有可能和数位dp结合一下.
四、组合数取模4：exLucas——模数p可以是合数的情况
p是合数的情况不好处理，
像之前学的CRT，如果p是合数，我们拆成若干素数幂的形式。
比如：
$ \left( \begin{matrix} n\\ m \end{matrix} \right) \tag{2} \bmod q$
\(q = p1^{e1}*p2^{e2}...pk^{ek}\)
$ \left( \begin{matrix} n\\ m \end{matrix} \right) \tag{2} \bmod pi^{ei}$
再用\(CRT\)可以解决
即：我们要解决
$ \left( \begin{matrix} n\\ m \end{matrix} \right) \tag{2} \bmod pi^{ei}$ \((p^e<=1e6)\)
\(CRT\)
\(m1,m2...mn\)
找到\(xi\)满足:\(xi \bmod mi = 1\)
$ xi \bmod \dfrac{M}{mi} = 0$;
\(x ≡ ai(\bmod mi)\)
\(x = Σxi*ai\)
\(x\)就是答案
\(Q:\)考虑为什么我们不能用之前的逆元的写法
\(A:\)逆元写法：
\(C_{n}^{m} = \dfrac{n!}{m!*(n-m)!}\)
\(\dfrac{a}{b} \bmod p\) ==> \(a \bmod p*(b \bmod p)^{-1}\)
即:
1.\(n!*(m!)^{-1}*(n-m)^{-1}\)
2.\(inv[i] = (p-\dfrac{p}{i})*inv[p\)%\(i]\)%\(p\)
1.因为n可能很大，直接算阶乘可能算不出来
2.算逆元的话，分母不一定是可逆的,比如\(p=2\)，分母有很多\(2\)的因子,所以说它不一定是可逆的
整体思路就是把阶乘都用\(n!= p^{an}*bn\)表示出来,其中p不整除\(bn\)(把\(p\)的因子都提出来,那这个\(bn\)相当于是多出来的)
\(n!= p^{an}*bn\)
\(C_{n}^{m} = \dfrac{p^{an}*bn}{p^{am}*bm*p^{an-m}*bn-m}\)
\(= \dfrac{p^{an-am-(an-m)}*bn}{bm*(bn-m)}\)
这里\(b\)是没有\(p\)的因子,肯定与\(p\)的幂次是互质的,那我们可以直接求其逆元
\(= p^{an-am-(an-m)}*bn*(bm*(bn-m))^{-1} (\bmod p^e)\)
画个表...
\(n!= p^{an}*bn\)
预处理：
\(fac[i]：1到i\)之间不被\(p\)整除%\(p^e\)的所有数的乘积
\(fac[p^e-1]^{n/p^e}*fac[n\)%\(p^e]\)
有\(\dfrac{n}{p^e}\)个完整的组，和n%\(p^e\)多出来的零头
递归下去就是：\(p^{\frac{n}{p}}*\dfrac{n}{p}!\)————>\(\dfrac{n}{p}!\)就是子问题
一直递归到\(<p\)为止
每个东西都算完之后我们用\(CRT\)合并
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 101000;
int m,T,M,phipe;
pair<int,int>x[110];
ll pr[110];
ll ans[N],a[N],b[N],fac[1010000];
ll ksm(ll a,ll b,ll m)
{
ll ans = 1,base = a;
while(b>0)
{
if(b&1)
{
ans *= base;
ans %= m;
}
base *= base;
base%=m;
b>>=1;
}
return ans;
}
pair<ll,ll>calc(ll a,int p,int pe)
{
ll cntp = 0,cnts = 0,val = 1;
while(a)
{
cntp += a/p;
cnts += a/pe;
val = val*fac[a%pe]%pe;
//val = val*ksm(fac[pe],a/pe)%pe*fac[a%pe]%pe
a/=p;
}
//val = val*ksm(fac[pe],cnts,pe)%pe;
val = val*ksm(fac[pe-1],cnts%phipe,pe)%pe;
return {val,cntp};//非素数部分和素数部分
}
ll binom(ll a,ll b,int p,int pe)
{
/*
n！= p^an*bn
Cnm = p^an*bn/(p^am*bm*p^(an-m)*(bn-m))
= p^(an-am-(an-m))*bn/(bm*(bn-m))
= p^(an-am-(an-m))*bn*(bm*(bn-m))^(-1) (mod p^e)
*/
auto f1 = calc(a,p,pe);
auto f2 = calc(b,p,pe);
auto f3 = calc(a-b,p,pe);
ll v1 = f1.first*ksm(f2.first*f3.first%pe,phipe-1,pe)%pe;//与p互质的部分
ll v2 = ksm(p,f1.second-f2.second-f3.second,pe);//与p不互质的部分
return v1*v2%pe;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cin>>m>>T;
M = m;
int t = 0;
for(int i = 2;i<=m;i++)
{
if(m%i==0)//对m进行分解
{
int p = i,pe = 1;
while(m%i==0)m/=i,pe*=i;
x[t++] = {p,pe};//x记录分解结果
}
}
/*
CRT
m1,m2...mn
找到xi满足:xi mod mi = 1
xi mod M/mi = 0;
x ≡ ai(mod mi)
x = Σxi*ai
*/
for(int i = 0;i<t;i++)
{
int pe = x[i].second;
for(int c = 0;c<M;c++)//这里M很小我们可以直接暴力for一遍找xi，就不用写exgcd了
{
if(c%pe==1&&c%(M/pe)==0)
{
pr[i] = c;
break;
}
}
}
for(int i = 0;i<T;i++) cin>>a[i]>>b[i];//先读进来再每个因子分开处理
for(int i = 0;i<t;i++)
{
int p = x[i].first,pe = x[i].second;
fac[0] = 1;//fac[i]:1~i之间不被p整除%p^e的所有数的乘积
for(int j = 1;j<=pe;j++)
{
if(j%p==0)
fac[j] = fac[j-1];
else
fac[j] = fac[j-1]*j%pe;//注意是%pe不是p
}
//phi(x) = x*∏(i=1~n)(1-1/pi)
//phipe = pe*(1-1/p);
phipe = (pe/p)*(p-1);
for(int j = 0;j<T;j++)
{
// b
//Ca mod (p^pe)
ans[j] = (ans[j]+binom(a[j],b[j],p,pe)*pr[i])%M;//用CRT并起来
}
}
for(int j = 0;j<T;j++)
cout<<ans[j]<<'\n';
return 0;
}
封装代码
#include<bits/stdc++.h>
using namespace std;
struct exLucas
{
const int N = 1e6+10;
typedef long long ll;
ll n,m,p,T;
inline ll ksm(ll a,ll b,const ll m)
{
ll ans = 1,base = a;
while(b>0)
{
if(b&1)
{
ans *= base;
ans %= m;
}
base *= base;
base%=m;
b>>=1;
}
return ans;
}
ll fac(const ll n,const ll p,const ll pe)
{
//pe:p^e
//fac[i]:1~i之间不被p整除%pe的所有数的乘积
if(!n)
return 1;
//ans包括了：n/(pe)个完整的组，和n%(pe)个零头
ll ans = 1;
//fac[pe-1]^(n/pe)*fac[n%pe]
for(int i = 1;i < pe;i++)
if(i%p)
ans = ans*i%pe;
ans = ksm(ans,n/pe,pe);
for (int i = 1; i <= n % pe; i++)
if (i % p)
ans = ans * i % pe;
return ans*fac(n/p,p,pe)%pe;//递归,分解成子问题
}
ll exgcd(const ll a,const ll b,ll &x,ll &y)
{
if(b==0)
{
x = 1,y = 0;
return a;
}
int d = exgcd(b,a%b,y,x);
y -= (a/b)*x;
return d;
}
ll inv(const ll a,const ll p)
{
ll x,y;
exgcd(a,p,x,y);
return (x%p+p)%p;
}
ll C(const ll n,const ll m,const ll p,const ll pe)
{
/*
Cnm
n！= p^an*bn
Cnm = p^an*bn/(p^am*bm*p^(an-m)*(bn-m))
= p^(an-am-(an-m))*bn/(bm*(bn-m))
= p^(an-am-(an-m))*bn*(bm*(bn-m))^(-1) (mod p^e)
*/
if (n < m)
return 0;
//Cnm = n!/m!*(n-m)!
ll f1 = fac(n, p, pe), f2 = fac(m, p, pe), f3 = fac(n - m, p, pe), cnt = 0;//计算与p互质的部分
//统计与p不互质的部分
for (ll i = n; i; i /= p)
cnt += i / p;
for (ll i = m; i; i /= p)
cnt -= i / p;
for (ll i = n - m; i; i /= p)
cnt -= i / p;
return f1 * inv(f2, pe) % pe * inv(f3, pe) % pe * ksm(p, cnt, pe) % pe;
}
ll a[N],c[N];
int cnt;
inline ll CRT()
{
/*
CRT
m1,m2...mn
找到xi满足:xi mod mi = 1
xi mod M/mi = 0;
x ≡ ai(mod mi)
x = Σxi*ai
*/
ll M = 1,ans = 0;
for(int i = 0;i<cnt;i++)
M *= c[i];
for(int i = 0;i<cnt;i++)
ans = (ans+a[i]*(M/c[i])%M*inv(M/c[i],c[i])%M)%M;
return ans;
}
ll exlucas(const ll n,const ll m,ll p)
{
cnt = 0;
ll tmp = sqrt(p);
for(int i = 2;i<=tmp&&p>1;i++)
{
ll tmp = 1;
while(p%i==0)
p/=i,tmp*=i;
if(tmp>1)
a[cnt] = C(n,m,i,tmp),c[cnt++] = tmp;
}
if(p>1)
a[cnt] = C(n,m,p,p),c[cnt++] = p;
return CRT();
}
int solve()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
cin >> n >> m >> p;
cout << exlucas(n, m, p)<<endl;
return 0;
}
};
