The number of solutions of Bzoj 3129 [Sdoi2013] equation and the modulus of combination number

Test Instructions:Link Method:The number of solutions of indefinite equation and the modulus of combination number parsing:First look at the limitations of N1 and N2 parts. For the latter part of the limit, we subtract directly from A n1+i ?1 You can convert a positive integer solution. But what about the first half? It's like a monkey. So we can do it. But this problem doesn't seem to be a good place to write. This is not a gift of TMD! How to take the model of large combination number? Please see my "Bzoj gift". In addition, the problem is x1+x2+...+xn=m is not less than equal Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N#define PA pair<int,int>#define MP Make_pair#define FI First#define SE SecondUsing namespace Std;typedef long long ll;ll a[n];ll t,p,tot;ll prime[n],mod[n],num[n],ans[n];void exgcd (ll a,ll B,ll &x, LL &y, LL &AMP;GCD) {if(!B) {x=1,y=0, Gcd=a;return; } EXGCD (B,a%b,y,x, GCD);y=y-a/b*x;} ll Get_inv (LL N,llm) {LLx,y, GCD; EXGCD (N,m,x,y, GCD);//(x/gcd%(m/GCD) +m/gcd)% (M/GCD)//gcd==1    return(x%m+m)%m;} ll Quick_my (LLx, LLy, ll MOD) {ll ret=1; while(y)    {if(y&1) ret= (ret*x)%mod;x=(x*x)%mod;y>>=1; }returnRET;} void Getp () {int x=p; for(intI=2; I*i<=x; i++) {if(x%i==0) {prime[++tot]=i,mod[tot]=1, num[tot]=0; while(x%i==0)            {x/=i,mod[tot]*=i,num[tot]++; }        }    }if(x!=1) {prime[++tot]=x, mod[tot]=x, num[tot]=1; }}ll CRT () {ll ret=0; for(intI=1; i<=tot;i++) {ret= (Ret+p/mod[i]*GET_INV(P/mod[i],mod[i])%p*ans[i])%p; }returnRET;} PA Get_factor (LLx, ll Pri,ll Mod) {ll ans=1;if(x==0||x==1)returnmp0,1); ll cnt_circle=x/mod,cnt_pri=x/pri;if(cnt_circle!=0)    { for(intI=2; i<mod;i++) {if(I%pri!=0) ans= (ans*(ll) i)%mod;    } ans=quick_my (Ans,cnt_circle,mod); } for(intI=cnt_circle*mod+1; i<=x; i++) {if(I%pri!=0) ans= (ans*(ll) i)%mod; } pa tmp=get_factor (Cnt_pri,pri,mod);returnMP (Cnt_pri+tmp.fi,ans*tmp. SE%mod);} ll Get_c (LL N,llm,intK) {if(m==0)return 1; PA tmp=get_factor (N,prime[k],mod[k]), Tmp1=get_factor (m, Prime[k],mod[k]), Tmp2=get_factor (nm, Prime[k],mod[k]);returnQuick_my (Prime[k],tmp.fi-tmp1.fi-tmp2.fi,mod[k])*tmp. SE%modK*GET_INV(Tmp1.se,mod[k])%modK*GET_INV(Tmp2.se,mod[k])%mod[K];} LL Calc (ll N,llm){if(n==m||m==0)return 1;if(n<m)return 0; for(intI=1; i<=tot;i++) Ans[i]=get_c (n,m, i);returnCRT ();} ll Ans;ll N,n1,n2,m; void Dfs (intNow,ll cnt,ll sum) {if(now==n1+1)    {if(CNT%2==0) ans= (Ans+calc (m-sum-1, N-1))%p;ElseAns= ((Ans-calc (m-sum-1, N-1))%p+P)%p;return; } DFS (now+1, cnt,sum); DFS (now+1, cnt+1, Sum+a[now]);}intMain () {scanf ("%lld%lld", &t,&p); GETP (); while(t--) {scanf ("%lld%lld%lld%lld",&n,&n1,&n2,&m); for(intI=1; i<=n1;i++) scanf ("%lld", &a[i]); for(intI=1; i<=n2;i++) {llx; scanf"%lld",&x);m=m-x+1; }if(m<=0) {puts ("0");Continue;} ans=0; Dfs1,0,0);printf("%lld\ n", ANS); }}

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The number of solutions of Bzoj 3129 [Sdoi2013] equation and the modulus of combination number