hdu--5396 (interval dp+ permutation combination)

To do this problem, think of whether it is DP, and then found that DP can do, but has been the pit to death.

enumerate the last merged position, and then for the minus, the two parts of the divided into a different combination of methods,
(A1+a2 ...) + (B1,b2 ............) For each A, the number of the added B factorial times, for each B, plus a number of factorial times
subtraction in the same vein
a special point of multiplication
(A1+a2 ...) * (B1,B2) * (...) * * * * ...) Multiply the allocation rate by multiplying the sum of the two parts directly

Think of these are far from the end, because the last two intervals merged, but also involved in a sequence of problems, a good start is silly and naïve directly by 2, the results of hehe.

If there are n symbols, I x on the left, and Y on the right, then because I have enumerated the last position of the merge, the last sign of the operation is determined, and after processing the number of the left and right numbers are added several times, the resulting total number is also processed because the order of merging is different, that is, C ( N-1,X) The relative order of the left-hand symbol in the X-position chosen is the same, because the previous order has been processed. I understand that there is a problem at the beginning, always mixed, especially afraid of DP count, especially with the combination of permutations, feeling the brain a bit worse ...

#pragma COMMENT (linker, "/stack:102400000,102400000") #include <cstdio> #include <cstring> #include < vector> #include <queue> #include <cmath> #include <string> #include <map> #include < utility> #include <algorithm> #include <iostream> #define MEM (b) memset (A,b,sizeof (a)) #define INF 0x3f3f3fusing namespace Std;int dx[]= { -1,0,1,0};///n e s Wint dy[]= {0,1,0,-1};typedef long long ll;typedef pair<int,i nt> pii;const int maxd=200+5;///==============================const int mod=1000000007;ll Dp[maxd][maxd],a[maxd],    Fac[maxd],c[maxd][maxd];int N;char str[maxd];void init () {fac[0]=1;    for (int i=1; i<maxd; i++) Fac[i]=fac[i-1]*i%mod;    MEM (c,0);        for (int i=0; i<maxd; 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;    }}ll DFS (int l,int r) {ll &res=dp[l][r];    if (res!=mod) return res;    if (l==r) return res=a[l];    res=0; for (int i=l; I&LT;r;        i++) {ll tmp;        if (str[i]== ' * ') Tmp=dfs (l,i) *dfs (i+1,r)%mod;        else if (str[i]== '-') tmp= (Dfs (l,i) *fac[r-i-1]-dfs (i+1,r) *fac[i-l]+mod)%mod;        else if (str[i]== ' + ') tmp= (Dfs (l,i) *fac[r-i-1]+dfs (i+1,r) *fac[i-l])%mod;    Res= (res+tmp*c[r-l-1][i-l]+mod)%mod; } return res;    int main () {int kase;    Freopen ("In.txt", "R", stdin);    Init ();        while (scanf ("%d", &n)!=eof) {for (int i=0; i<n; i++) scanf ("%lld", &a[i]);        scanf ("%s", str);        for (int i=0, i<n; i++) for (int j=0; j<n; j + +) Dp[i][j]=mod;    cout<< (Dfs (0,n-1) +mod)%mod<<endl; } return 0;} /*33 2 1-+51 4 6 8 3+*-*51 1 1 1 1*****/

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hdu--5396 (interval dp+ permutation combination)