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  1. #include <bits/stdc++.h> 
  2. using namespace std;  
  3.  
  4. typedef long long ll;  
  5. typedef pair<int, int> ii;  
  6.  
  7. const int INF = 1e9;  
  8. const ll LINF = 1e18;  
  9.  
  10. const int N = 1e2 + 5; 
  11. const int K = 1e5 + 5; 
  12. const int MOD = 1e9 + 7;  
  13.  
  14. int n, k;  
  15. int a[N]; 
  16. int dp[N][K]; // dp[i][j] = xét đến đứa trẻ thứ i và chia j cây kẹo 
  17. int sum[N][K]; // sum[i][j] = tổng các dp[i][1] + dp[i][2] + ... + dp[i][j]
  18.  
  19. int add(int a, int b) {
  20. 	int ans = (a + b) % MOD;
  21. 	if (ans < 0) ans += MOD; 
  22. 	return ans;  
  23. }
  24.  
  25. int getSum(int i, int lj, int rj) {
  26. 	int ans = sum[i][rj]; 
  27. 	if (lj - 1 >= 0) ans = add(ans, -sum[i][lj - 1]);  
  28. 	return ans;  
  29. }
  30.  
  31. int main() {
  32. 	ios::sync_with_stdio(0); cin.tie(0);  	
  33. 	cin >> n >> k; 
  34. 	for (int i = 1; i <= n; i++) cin >> a[i];  
  35.  
  36. 	dp[0][0] = 1;   
  37. 	for (int j = 0; j <= k; j++) sum[0][j] = 1; 
  38.  
  39. 	for (int i = 1; i <= n; i++) {
  40. 		for (int j = 0; j <= k; j++) {
  41. 			// for (int x = 0; x <= a[i]; x++) {
  42. 			// 	if (j - x >= 0) add(dp[i][j], dp[i - 1][j - x]); 
  43. 			// }
  44. 			dp[i][j] = getSum(i - 1, j - a[i], j); 
  45. 			sum[i][j] = dp[i][j]; 
  46. 		}
  47.  
  48.  		for (int j = 1; j <= k; j++) sum[i][j] = add(sum[i][j], sum[i][j - 1]); 
  49. 	}
  50.  
  51. 	cout << dp[n][k] << '\n'; 
  52. }
Success #stdin #stdout 0.01s 11304KB
comments (?)
stdin 
4 100000
100000 100000 100000 100000
stdout 
665683269
https://ideone.com/4sM1lH
language:
C++14 (gcc 8.3)
created:
1 year ago
visibility:
secret

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