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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. typedef long long LL;
  6. const int N = (int) 1e6 + 4005;
  7. const LL mod = (LL) 1e9 + 7;
  8.  
  9. LL c[N], fact[N], inv[N], dp[2005][2005], ans[4005];
  10.  
  11. // calculates (-1)^n
  12. int sign (int n) {
  13. 	if (n & 1) return -1;
  14. 	else return 1;
  15. }
  16.  
  17. // calculates a^b % mod
  18. LL modpow (LL a, LL b, LL mod) {
  19. 	LL res = 1;
  20. 	while (b) {
  21. 		if (b & 1) res = (res * a) % mod;
  22. 		a = (a * a) % mod;
  23. 		b >>= 1;
  24. 	}
  25. 	return res;
  26. }
  27.  
  28. // calculates nCr.
  29. LL C (LL n, LL r) {
  30. 	if (r < 0 || r > n) return 0;
  31. 	LL ans = (fact[n] * inv[n - r]) % mod;
  32. 	return (ans * inv[r]) % mod;
  33. }
  34.  
  35.  
  36. int main () {
  37. 	fact[0] = 1;
  38. 	// calculate factorial and inverse modulo p for each factorial.
  39. 	for (LL i = 0; i < N; i++) fact[i + 1] = (LL) (fact[i] * (i + 1)) % mod;
  40. 	for (int i = 0; i < N; i++) inv[i] = modpow (fact[i], mod - 2, mod);
  41.  
  42. 	// process the input and calculate array c_i as shown in the editorial.
  43. 	int n;
  44. 	scanf ("%d", &n);
  45. 	for (int i = 0; i < n; i++) {
  46. 		int foo;
  47. 		scanf ("%d", &foo);
  48. 		c[foo + 1] ++;
  49. 	}
  50.  
  51. 	// Now we come for the dp step.
  52. 	// dp[i][s] = as given in the editorial, coefficient of x^s in first i steps.
  53. 	dp[0][0] = 1;
  54. 	for (int i = 1; i <= 1001; i++) {
  55. 		for (int s = 0; s <= 2000; s++) {
  56. 			int k = i;
  57. 			LL sum = 0;
  58. 			// try all coefficients of x^(k * l), this is crucial step in complexity reduction.
  59. 			for (int l = 0; l <= s; l++) {
  60. 				if (k * l > s) break;
  61. 				// coefficient of x^(k * l) in (x^(i + 1) - 1) ^ c_i
  62. 				LL prod = C (c[i], l) * sign (c[i] - l);
  63. 				sum = (sum + dp[i - 1][s - k * l] * prod) % mod;
  64. 				if (sum < 0) sum += mod;
  65. 			}
  66. 			dp[i][s] = sum;
  67. 		}
  68. 	}
  69.  
  70. 	// Now do the multiplication part as shown in the editorial.
  71. 	for (int i = 0; i <= 2000; i++) {
  72. 		for (int j = 0; j <= 2000; j++) {
  73. 			// coefficients of x^j in (x - 1)^(-n) is (-1)^n C (n + j - 1, j), See the given link for explantion.
  74. 			ans[i + j] = (ans[i + j] + dp[1001][i] *  sign (n) * C (n + j - 1, j)) % mod;
  75. 			if (ans[i + j] < 0) ans[i + j] += mod; 
  76. 		}
  77. 	}
  78.  
  79. 	// finally take the query input as given in the problem.
  80. 	int Q;
  81. 	scanf ("%d", &Q);
  82. 	while (Q--) {
  83. 		int s;
  84. 		scanf ("%d", &s);
  85. 		printf ("%lld\n", ans[s]);
  86. 	}
  87.  
  88. 	return 0;
  89. }
Runtime error #stdin #stdout 2.63s 58272KB
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https://ideone.com/Uqqq62
language:
C++14 (gcc 8.3)
created:
10 years ago
visibility:
public

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