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1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 using namespace std; 6 #define MAXN 20 7 int n,T,m,p; 8 int a[20]; 9 int factorial[100001],rev[100001]; 10 void init() 11 { 12 int i; 13 factorial[0]=1; 14 for(i=1;i<p;i++) 15 factorial[i]=(long long)factorial[i-1]*i%p; 16 rev[1]=1; 17 for(i=2;i<p;i++) 18 rev[i]=((-(p/i)*(long long)rev[p%i])%p+p)%p; 19 } 20 int calc_p(int x) 21 { 22 int ret=0; 23 while(x) 24 { 25 ret+=x/p; x/=p; 26 } 27 return ret; 28 } 29 int my_pow(int x,int n) 30 { 31 int ans=1; 32 while(n) 33 { 34 if(n&1) ans=(long long)ans*x%p; 35 x=(long long)x*x%p; 36 n>>=1; 37 } 38 return ans; 39 } 40 int calc(int x) 41 { 42 int ans=1,ret=0; 43 while(x) 44 { 45 ret+=x/p; ans=(long long)ans*factorial[x%p]%p; 46 x/=p; 47 } 48 ans=(long long)my_pow(factorial[p-1],ret)*ans%p; 49 return ans; 50 } 51 52 int combination(int n,int m) 53 { 54 if(n<m) return 0; 55 if(calc_p(n)!=calc_p(m)+calc_p(n-m)) 56 return 0; 57 int s1=calc(n),s2=(long long)calc(m)*calc(n-m)%p; 58 return (long long)s1*rev[s2]%p; 59 } 60 int solve() 61 { 62 int ans=0; 63 m=m+n+1; 64 n++; 65 int top=(1<<T),i,j,cc,x; 66 for(i=0;i<top;i++) 67 { 68 cc=0; 69 x=m; 70 for(j=0;j<T;j++) 71 if(i&(1<<j)) 72 { 73 cc++; 74 x-=a[j]; 75 } 76 if(cc%2==0) ans+=combination(x-1,n-1); 77 else ans-=combination(x-1,n-1); 78 ans=(ans%p+p)%p; 79 } 80 return ans; 81 } 82 83 int main() 84 { 85 scanf("%d%d%d%d",&n,&T,&m,&p); 86 init(); 87 int i; 88 for(i=0;i<T;i++) scanf("%d",a+i), a[i]++; 89 printf("%d\n",solve()); 90 return 0; 91 92 }
思路：T很小 容斥原理