hdu 5396 Expression 区间DP+排列组合 2015 Multi-University Training Contest 9

Expression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 453    Accepted Submission(s): 264

Problem Description

Teacher Mai has n numbers a1,a2,⋯,anand n−1 operators("+", "-" or "*")op1,op2,⋯,opn−1, which are arranged in the form a1 op1 a2 op2 a3 ⋯ an.

He wants to erase numbers one by one. In i-th round, there are n+1−i numbers remained. He can erase two adjacent numbers and the operator between them, and then put a new number (derived from this one operation) in this position. After n−1 rounds, there is the only one number remained. The result of this sequence of operations is the last number remained.


He wants to know the sum of results of all different sequences of operations. Two sequences of operations are considered different if and only if in one round he chooses different numbers.

For example, a possible sequence of operations for "1+4∗6−8∗3" is 1+4∗6−8∗3→1+4∗(−2)∗3→1+(−8)∗3→(−7)∗3→−21.

 

Input

There are multiple test cases.

For each test case, the first line contains one number n(2≤n≤100).

The second line contains n integers a1,a2,⋯,an(0≤ai≤109).

The third line contains a string with length n−1 consisting "+","-" and "*", which represents the operator sequence.

 

Output

For each test case print the answer modulo 109+7.

 

Sample Input

33 2 1-+51 4 6 8 3+*-*

 

Sample Output

2999999689
Hint
 Two numbers are considered different when they are in different positions.
 

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

 

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区间DP，

dp[i][j]表示区间[i,j]所有可能的计算顺序之后结果的和。

考虑区间i,j最后一个计算的符号位置:设符号左边所有可能的计算顺序之后结果有a1,a2,...,an,右边可能的结果有b1,b2,b3,...,bn。（可重）

计算方法：首先将所有ai 根据最后一个符号和所有bi，+或-或*，然后再对其排列，因为每种ai (+或-或*) bi

左区间和右区间内的符号运行相对顺序可以不同，但是结果相同。

（左右区间自身的顺序是定的，因为用dp之前已算出子区间所有顺序的结果和）

'+':

dp[i][j]=C （组合公式） *  ( ∑ai * A( 右区间符号数全排列)   + ∑bi *A(左区间符号数全排列)  )

      （左右区间的相对顺序） （ 对于每个ai，一共加了A( 右区间符号数全排列)多次。）

dp[i][j]=(dp[i][k-1]*A( 右区间符号数全排列)+dp[k][i]*A(左区间符号数全排列))*C( , );      (k=i+1,...,j);

'-':dp[i][j]=(dp[i][k-1]*A( 右区间符号数全排列)-dp[k][i]*A(左区间符号数全排列))*C( , ); 

'*'::dp[i][j]=(dp[i][k-1]*dp[k][i])*C( , ); 

d[i][j]=C( ,)  *(a1*b1+a1*b2+...+a1*bn+a2*b1+...an*bn)=C( ,)  *∑ai*∑bi

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI 3.1415926535897932384626#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   num<<1,le,mid#define rson    num<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mp    make_pair#define _f     first#define _s     secondusing namespace std;const int INF =0x3f3f3f3f;const int maxn= 100+5   ;//const int INF=    ;typedef long long ll;const ll mod= 1000000000+7    ;const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);//by yskysker123

 

int n;int a[maxn];char c[maxn];ll dp[maxn][maxn];ll A[maxn];ll C[maxn][maxn];<span style="font-family: 'Times New Roman';">  //杨辉三角预处理全排列和组合结果</span>void pre(){    A[0]=1;    for(ll i=0;i<maxn;i++)       //C(0,x),C( x,0)的情况不能漏，这里可以视为1              {       if(i>0) A[i]=A[i-1]*i%mod;        for(ll j=0;j<=i;j++)                                {            if(j==0||j==i)  C[i][j]=1;                          else            {                C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;             }        }    }}int main(){    pre();    while(~scanf("%d",&n))    {        FOR0(i,n)        scanf("%d",&a[i]);        FOR1(i,n-1)        scanf(" %c",&c[i]);        memset(dp,0,sizeof dp);        for(int i=0;i<=n-1;i++)        {            dp[i][i]=a[i];        }        for(int j=0;j<=n-1;j++)  //计算的顺序        {            for(int i=j-1;i>=0;i--)            {                ll ret;                for(int k=i+1;k<=j;k++)                {                    if(c[k]=='*')                    {                        ret= (dp[i][k-1] * dp[k][j])%mod;                    }                    else if(c[k]=='+')                    {                        ret=(dp[i][k-1]*A[j-k  ]%mod +dp[k][j]*A[k-1-i]%mod )%mod;                    }                    else                    {                        ret=((dp[i][k-1]*A[j-k  ]%mod -dp[k][j]*A[k-1-i]%mod )+mod)%mod;                    }                    ret=ret*C[j-i-1][k-i-1 ]%mod;                     dp[i][j]=(dp[i][j]+ret)%mod;                }            }        }        printf("%lld\n",dp[0][n-1]);    }    return 0;}

本题用以下的组合公式不仅超时，而且是错的，因为取了模。

//ll C(ll n,ll m)//{//    ll ans=1;//    m=min(m,n-m);//    for(ll i=1;i<=m;i++)//    {//        ans=ans*(n-i+1)/i   %mod;////    }////    return ans;////}