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  1. // Ancient Prophecy (D), by Errichto
  2. // AC, O(n^2)
  3. #include<bits/stdc++.h>
  4. using namespace std;
  5. #define FOR(i,a,b) for(int i = (a); i <= (b); ++i)
  6. #define RI(i,n) FOR(i,1,(n))
  7. #define REP(i,n) FOR(i,0,(n)-1)
  8.  
  9. const int nax = 5005;
  10. const int mod = 1e9 + 7;
  11. const int inf = 1e9 + 120;
  12.  
  13. int t[nax];
  14. char sl[nax];
  15. // int dp[nax][nax];
  16. // pref[a][b] = sum(dp[1..a][b])
  17. int pref[nax][nax];
  18. int nxt[nax][nax];
  19.  
  20. int cmp(int i, int j) {
  21. 	if(t[i] < t[j]) return -1;
  22. 	if(t[i] > t[j]) return 1;
  23. 	return 0;
  24. }
  25.  
  26. int dp(int a, int b) {
  27. 	int x = pref[a][b] - pref[a-1][b];
  28. 	if(x < 0) x += mod;
  29. 	return x;
  30. }
  31.  
  32. int main() {
  33. 	int n;
  34. 	scanf("%d", &n);
  35. 	scanf("%s", sl);
  36. 	RI(i, n) t[i] = sl[i-1] - '0';
  37.  
  38. 	for(int a = n; a >= 1; --a)
  39. 		for(int b = n; b > a; --b) {
  40. 			int & x = nxt[a][b];
  41. 			if(t[a] != t[b]) x = a;
  42. 			else if(b == n) x = inf;
  43. 			else x = nxt[a+1][b+1];
  44. 		}
  45.  
  46. 	RI(i, n) {
  47. 		// dp[1][i] = 1;
  48. 		RI(j, n) pref[j][i] = 1;
  49. 	}
  50. 	FOR(c, 2, n) FOR(d, c, n) {
  51. 		int & x = pref[c][d];
  52. 		int b = c - 1;
  53. 		int a = b - (d - c);
  54. 		x = pref[b][b] - pref[max(a,0)][b];
  55. 		if(x < 0) x += mod;
  56. 		if(a >= 1) {
  57. 			int where = nxt[a][c];
  58. 			if(where < c && t[where] < t[c+where-a]) {
  59. 				x += dp(a, b);
  60. 				if(x >= mod) x -= mod;
  61. 			}
  62. 		}
  63. 		if(t[c] == 0) x = 0;
  64. 		pref[c][d] = x + pref[c-1][d];
  65. 		if(pref[c][d] >= mod) pref[c][d] -= mod;
  66. 	}
  67. 	int ans = 0;
  68. 	RI(i, n) ans = (ans + dp(i, n)) % mod;
  69. 	printf("%d\n", ans);
  70. 	return 0;
  71. }
Runtime error #stdin #stdout 0s 199168KB
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https://ideone.com/zczsc3
D
language:
C++14 (gcc 8.3)
created:
8 years ago
visibility:
secret

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