51NOD1020 reverse order (DP)

1020 reverse permutation base time limit: 2 seconds Space limit: 131072 KB score: 80 Difficulty: 5-level algorithm topic collection focus on a permutation, if the number of the front and back position of the opposite of the size order, that is, the previous number is greater than the next number, then they are called a reverse order. The total number of reverse order in a permutation is called the inverse number of the permutation. such as 2 4 3 1, 2 1,4 3,4 1,3 1 is reverse, the reverse number is 4.
1-n in the whole arrangement, the number of reverse order is 0 (positive), the maximum is n (n-1)/2 (reverse) given 2 numbers n and K, for the 1-n of the whole arrangement, the number of reverse order is k of how many kinds? For example: n = 4 K = 3.
1 2 3 4 A total of 3 in reverse order 6, respectively: 1 4 3 22 3 4 12 4 1 33 1 4 23 2 1 44 1 2 3
Since the number of reverse permutations is very large, it is only possible to calculate and output the number of Mod 10^9 + 7 results. Input

Line 1th: A number t that represents the number of numbers that are later used as input tests. (1 <= T <= 10000) 2-t + 1 lines: 2 numbers per line n,k. The middle is separated by a space. (2 <= N <=, 0 <= k <= 20000)

Output

Total T-line, corresponding to the number of reverse order Mod (10^9 + 7)

Input example

14 3

Output example

6

Set F (n,k) indicates the number of permutations in the order of N of the number of numbers in reverse order of K.

The largest number of n may be ranked in the first n-i, resulting in an I and N-related reverse order, minus N, the remaining number of n-1 is k-i in reverse order. So, f (n,k) = SUM (f (n-1,k-i)) (0<=i<n). Similarly there is f (n,k-1) = SUM (f (n-1,k-1-i)) (0<=i<n). Subtract from the two, you can get F (n,k)-F (n,k-1) =f (N-1,k)-F (n-1,k-n). The recursive formula is F (n,k) =f (n,k-1) +f (N-1,k)-F (n-1,k-n).

#include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <string > #include <functional> #include <cmath> #include <set> #include <queue> #include <  algorithm> #include <vector> #include <map> #include <stack>using namespace std; #define ESP 1e-8const Double PI = ACOs ( -1.0); const double e = 2.718281828459;const int inf = 2147483647;const long Long mod = 10000000 07;typedef Long Long Ll;//freopen ("In.txt", "R", stdin); input redirection, input data will be read from the In.txt file//freopen ("OUT.txt", "w", stdout);  Output redirection, the output data will be saved in the OUT.txt file int dp[1005][20005];void init () {for (int i = 1; I <=, ++i) dp[i][0] = 1;for (int i = 2; I <= 1000; ++i) {for (int j = 1; J <= I * (i-1)/2 && J <= 20000; ++j) {Dp[i][j] = (Dp[i][j-1] + dp[i-1][j])% mod if (j-i >= 0) dp[i][j] = dp[i-1][j-i];DP [i][j] = (dp[i][j]% mod + mod)% mod;}} int main () {int T, N, K;init (), scanf ("%d", &t), while (t--) {scanf ("%d%d", &n, &k);p rintf ("%d\n", Dp[n][k]);}} 

51NOD1020 reverse order (DP)