HDU 5396 interval DP Math Expression

Test instructions: There are n numbers, n-1 operators, the order of each operator can be arbitrary, so there is a total (n-1)! Order of operations, get (N-1)! The results of these operations and the MOD 1e9+7.

Analysis:

The position K of the last operator in an enumeration interval [L, R], which is an analogy to the optimal matrix chain multiplication.

If the operator is multiplicative, then the multiplication is assigned according to the multiplication allocation rate.

There is a total of r-l operators in this interval, where the last operator has been determined to be K, the left interval [L, K] has the K-L operator, the right interval [k + 1, R] has r-k-1 operators.

Moreover, the sequence of the left and right interval operators is determined, and the order between the two intervals is not affected, so the same results are in total C (R-l-1, K-L)

So the answer will be multiplied by this number, D (i, j) + = d (i, K) * d (k + 1, R) * C (R-l-1, L-k) | OP[K] = *

However, if you add and subtract, you cannot directly follow the operator to merge the interval.

For the determination of the left interval a sequence of operations, the right interval has a total (R-k-1)! Result of the operation, so the answer accumulates a D (l, K) * (R-K-1)!

Similarly, for the right interval a definite sequence of operations, the left interval pair should (K-l)! Result of the operation, the answer is summed up a D (k + 1, R) * (L)!

Finally, the order of two interval r-l-1 operators is determined, and the final answer is multiplied by C (R-l-1, K-L)

A final summary of the answer is:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 using namespacestd;6 7typedefLong LongLL;8 9 Const intMAXN = -+Ten;Ten ConstLL MOD =1000000007; One  A intN; - LL A[MAXN]; - LL FAC[MAXN], C[MAXN][MAXN]; the CharOP[MAXN]; -  - intVIS[MAXN][MAXN]; - LL D[MAXN][MAXN]; +  -LL DP (intLintR) + { A     if(Vis[l][r])returnD[l][r]; atll& ans =D[l][r]; -Ans =0; -VIS[L][R] =true; -     if(L = = r)returnAns =A[l]; -     if(L +1==R) -     { in         if(Op[l] = ='*')returnAns = a[l] * A[r]%MOD; -         if(Op[l] = ='+')returnAns = (A[l] + a[r])%MOD; to         if(Op[l] = ='-')returnAns = (((((A[l]-a[r])% MoD) + MoD)%MOD; +     } -      for(intK = l; K < R; k++) the     { *LL T1 = DP (l, k), t2 = DP (k +1, R); $ LL t;Panax Notoginseng         if(Op[k] = ='*') -         { thet = T1 * T2%MOD; +t = T * c[r-l-1][k-l]; AAns = (ans + t)%MOD; the             Continue; +         } -  $T1 = T1 * Fac[r-k-1] %MOD; $t2 = T2 * Fac[k-l]%MOD; -         if(Op[k] = ='+') T = (t1 + t2)%MOD; -         ElseT = ((((t1-t2)% MoD) + MoD)%MOD; thet = T * c[r-l-1][k-l]; -Ans = (ans + t)%MOD;Wuyi     } the  -     returnans; Wu } -  About intMain () $ { -fac[0] =1; -      for(inti =1; i < MAXN; i++) Fac[i] = fac[i-1] * I%MOD; -      for(inti =0; i < MAXN; i++) c[i][0] = C[i][i] =1LL; A      for(inti =2; i < MAXN; i++) +          for(intj =1; J < I; J + +) C[i][j] = (c[i-1][J] + c[i-1][j-1]) %MOD; the  -      while(SCANF ("%d", &n) = =1&&N) $     { the          for(inti =1; I <= N; i++) scanf ("%i64d", A +i); thescanf"%s", OP +1); thememset (Vis,false,sizeof(Vis)); thememset (Vis,0,sizeof(Vis)); -printf"%i64d\n", DP (1, N)); in     } the  the     return 0; About}

code June

HDU 5396 interval DP Math Expression