HDU-5155 Harry and Magic Box

Problem description
A box is made up of n*m, and there are some glittering gems in the lattice.
Now there is a box from the left to see the past, each line is shining, from the front to see the past, each column is also flashing light.
Q: How many types of gems are in the box?
The answer is likely to be large, so the output answer is 1000000007 modulo.


Enter a description
Multiple sets of input data
One row per set of data, enter two number n m for the size of the box, 0≤n,m≤50


Output description
Each set of data outputs a row, an integer, representing the number of scenarios


Analytical:
This question looks best.
Coder above the puzzle, really is to see the dazed, and later asked a little seniors, just understand how to do.


The main use of the idea of recursion.
Draw a diagram to explain:
Scenario 1:



Scenario 2:



In both cases, the state transfer formula is obtained:

DP[I][J] + = dp[i][j-1]* (2^i-1) + dp[i-k][j-1] * c[k][i] * (2^ (i-k))

#include <cstdio> #include <cstring> #include <cmath> #include <algorithm>using namespace std; typedef __int64 LL;CONST int MOD = 1000000007;const int N = 51;ll c[n][n],dp[n][n],t[n];int n,m;void init () {memset (c,0,si Zeof (c)); C[0][0] = 1;for (int i = 1;i <= N; i++) {c[i][0] = 1;for (int j = 1;j <= i; j + +) C[i][j] = (C[i-1][j-1] + c[ I-1][J])% MOD;} T[0] = 1;for (int i = 1; I <= N; i++) {t[i] = t[i-1] *% MOD;}}  int main () {init (); while (scanf ("%d%d", &n,&m)! = EOF) {for (int i = 0; I <= N; i++) dp[i][1] = 1;for (int j = 0; j <= m; J + +) Dp[1][j] = 1;for (int i = 2; I <= n; i++) {for (int j = 2; J <= M; j + +) {Dp[i][j] = dp[i][j-1] * (T[i]-1)% MOD; for (int k = 1; k < i; k++) {Dp[i][j] = (Dp[i][j] + dp[i-k][j-1] * c[i][k]% mod * t[i-k]% mod)% mod;}} printf ("%i64d\n", Dp[n][m]);} return 0;}

HDU-5155 Harry and Magic Box