两侧同时换到之前的修订记录 前一修订版 后一修订版 | 前一修订版 | ||
2020-2021:teams:namespace:牛客多校第六场 [2020/07/30 17:31] great_designer [F] |
2020-2021:teams:namespace:牛客多校第六场 [2020/08/01 10:22] (当前版本) great_designer [I] |
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=====I===== | =====I===== | ||
- | [[Stling数]] | + | 题目、解答与代码见[[Stirling数]]。 |
- | + | ||
- | <hidden> | + | |
- | <code C> | + | |
- | + | ||
- | #include<stdio.h> | + | |
- | + | ||
- | int MOD; | + | |
- | + | ||
- | long long powmod(long long x,int k) | + | |
- | { | + | |
- | long long ans=1; | + | |
- | while(k) | + | |
- | { | + | |
- | if(k&1) | + | |
- | { | + | |
- | ans=ans*x%MOD; | + | |
- | } | + | |
- | x=x*x%MOD; | + | |
- | k>>=1; | + | |
- | } | + | |
- | return ans; | + | |
- | } | + | |
- | + | ||
- | int getfact(int x,int *p) | + | |
- | { | + | |
- | int t=x,sz=0; | + | |
- | int i; | + | |
- | for(i=2;i*i<=t;i++) | + | |
- | { | + | |
- | if(x%i==0) | + | |
- | { | + | |
- | p[++sz]=i; | + | |
- | while(x%i==0) | + | |
- | { | + | |
- | x/=i; | + | |
- | } | + | |
- | } | + | |
- | } | + | |
- | if(x>1) | + | |
- | { | + | |
- | p[++sz]=x; | + | |
- | } | + | |
- | return sz; | + | |
- | } | + | |
- | + | ||
- | long long facd[1000005],facv[1000005]; | + | |
- | long long G,mi[1000005],inv[1000005]; | + | |
- | + | ||
- | void pre() | + | |
- | { | + | |
- | facd[0]=1; | + | |
- | int i; | + | |
- | for(i=1;i<MOD;i++) | + | |
- | { | + | |
- | facd[i]=facd[i-1]*i%MOD; | + | |
- | } | + | |
- | facv[MOD-1]=facd[MOD-1]; | + | |
- | for(i=MOD-2;i>=0;i--) | + | |
- | { | + | |
- | facv[i]=facv[i+1]*(i+1)%MOD; | + | |
- | } | + | |
- | int prime[10]; | + | |
- | int sz=getfact(MOD-1,prime); | + | |
- | for(G=1;;G++) | + | |
- | { | + | |
- | bool ok=1; | + | |
- | for(i=1;i<=sz;i++) | + | |
- | { | + | |
- | if(powmod(G,(MOD-1)/prime[i])==1) | + | |
- | { | + | |
- | ok=0; | + | |
- | break; | + | |
- | } | + | |
- | } | + | |
- | if(ok) | + | |
- | { | + | |
- | break; | + | |
- | } | + | |
- | } | + | |
- | mi[0]=1; | + | |
- | for(i=1;i<MOD-1;i++) | + | |
- | { | + | |
- | mi[i]=mi[i-1]*G%MOD; | + | |
- | } | + | |
- | inv[1]=1; | + | |
- | for(i=2;i<MOD;i++) | + | |
- | { | + | |
- | inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD; | + | |
- | } | + | |
- | } | + | |
- | + | ||
- | long long C(long long n,long long m) | + | |
- | { | + | |
- | return (n<m)?0:facd[n]*facv[m]%MOD*facv[n-m]%MOD; | + | |
- | } | + | |
- | + | ||
- | long long calc(long long n,long long m) | + | |
- | { | + | |
- | if(!m) | + | |
- | { | + | |
- | return 1; | + | |
- | } | + | |
- | if(n<m) | + | |
- | { | + | |
- | return 0; | + | |
- | } | + | |
- | return C(n%MOD,m%MOD)*calc(n/MOD,m/MOD)%MOD; | + | |
- | } | + | |
- | + | ||
- | long long query(long long n,long long m) | + | |
- | { | + | |
- | m=(n<m)?n:m; | + | |
- | long long s=0; | + | |
- | int i; | + | |
- | for(i=0;i<MOD-1;i++) | + | |
- | { | + | |
- | int x=mi[i]; | + | |
- | if(x+n>MOD) | + | |
- | { | + | |
- | continue; | + | |
- | } | + | |
- | if(!i) | + | |
- | { | + | |
- | s=(s+(m+1)*facd[x+n-1]%MOD*facv[x-1])%MOD; | + | |
- | } | + | |
- | else | + | |
- | { | + | |
- | s=(s+(mi[(MOD-1-i*(m+1)%(MOD-1))%(MOD-1)]-1LL)*inv[mi[MOD-1-i]-1LL]%MOD*facd[x+n-1]%MOD*facv[x-1])%MOD; | + | |
- | } | + | |
- | } | + | |
- | s=(MOD-s)%MOD; | + | |
- | if(n==MOD-1) | + | |
- | { | + | |
- | s=(s-1LL+MOD)%MOD; | + | |
- | } | + | |
- | return s; | + | |
- | } | + | |
- | + | ||
- | long long solve(long long n,long long m) | + | |
- | { | + | |
- | long long u1=n/MOD,v1=n%MOD; | + | |
- | if(m<u1) | + | |
- | { | + | |
- | return 0; | + | |
- | } | + | |
- | m-=u1; | + | |
- | long long u2=m/(MOD-1),v2=m%(MOD-1); | + | |
- | long long s=0; | + | |
- | if(u2) | + | |
- | { | + | |
- | s=(s+(((u2-1)&1)?MOD-1:1)*calc(u1-1,u2-1)%MOD*query(v1,MOD-2))%MOD; | + | |
- | } | + | |
- | s=(s+((u2&1)?MOD-1:1)*calc(u1,u2)%MOD*query(v1,v2))%MOD; | + | |
- | if(v1==MOD-1&&u2) | + | |
- | { | + | |
- | s=(s+(((u2-1)&1)?MOD-1:1)*calc(u1-1,u2-1))%MOD; | + | |
- | } | + | |
- | return s*((u1&1)?MOD-1:1)%MOD; | + | |
- | } | + | |
- | + | ||
- | int main() | + | |
- | { | + | |
- | long long n,l,r; | + | |
- | scanf("%lld%lld%lld%d",&n,&l,&r,&MOD); | + | |
- | pre(); | + | |
- | long long s1=solve(n,r); | + | |
- | long long s2=solve(n,l-1); | + | |
- | printf("%lld\n",(s1-s2+MOD)%MOD); | + | |
- | return 0; | + | |
- | } | + | |
- | + | ||
- | </code> | + | |
- | </hidden> | + | |
=====K===== | =====K===== |