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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define mod 1000000007
  4. long long a[1000009]={0},b[1000009]={0};
  5. long long invp(int a,int n)
  6. {
  7. 	if(a==1 || n==0)return 1;
  8. 	if(n==1)return a;
  9. 	long long t=invp(a,n/2);
  10. 	t=(t*t)%mod;
  11. 	if(n&1)
  12. 	t=(t*a)%mod;
  13. 	return t;
  14. }
  15. void initnum()
  16. {
  17. 	a[1]=a[0]=1;
  18. 	b[0]=b[1]=1;
  19. 	for(int i=2;i<=1000006;i++)
  20. 		a[i]=(a[i-1]*i)%mod;
  21. 	b[1000007]=1;
  22. 	for(int i=2;i<=1000007;i++)
  23. 		b[1000007]=(b[1000007]*invp(i,mod-2))%mod;
  24. 	for(int i=1000006;i>=2;i--)
  25. 		b[i]=(b[i+1]*(i+1))%mod;
  26. }
  27.  
  28. long long ncr(int n,int r)
  29. {
  30. 	return (((a[n]*b[r])%mod)*b[n-r])%mod;
  31. }
  32. int main() 
  33. {
  34. 	initnum();
  35. //	initdenum();
  36. 	int k;
  37. 	cin>>k;
  38. 	int c[1009]={0};
  39. 	for(int i=1;i<=k;i++)cin>>c[i];
  40. 	long long dp[1009]={0};
  41. 	dp[1]=1;
  42. 	int sum=c[1];
  43. 	for(int i=2;i<=k;i++)
  44. 	{
  45. 		dp[i]=dp[i-1];
  46. 		if(c[i]>1)dp[i]=(dp[i]*ncr(sum+c[i]-1,c[i]-1))%mod;
  47. 		sum+=c[i];
  48. 	}
  49. 	cout<<dp[k]<<endl;
  50. 	return 0;
  51. }
Success #stdin #stdout 1.92s 18768KB
comments (?)
stdin 
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stdout 
1680
https://ideone.com/3FLh1G
language:
C++ (gcc 8.3)
created:
9 years ago
visibility:
public

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