hdu5396 Expression

Expression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 952    Accepted Submission(s): 573

Problem Description

Teacher Mai has n numbers a1,a2,⋯,an and 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≤10^9) .

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 10^9+7 .

 

Sample Input

3

3 2 1

-+

5

1 4 6 8 3

+*-*

Sample Output

2

999999689

Hint

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

一看这样子就像是区间dp

再看看数据肯定是区间dp

f[i][j]表示区间[i,j]一共(j-i)!种运算得到的所有数之和

加减都很简单，枚举区间[i,j]中i到j-1中间最后一个运算符k

如果是加号，f[i][j]+=(f[i][k]*(j-k-1)!+f[k+1][j]*(k-i)!)*C(j-i-1,k-i)

意思就是考虑左右两边的f[i][k],f[k+1][j]对f[i][j]的影响

如果是减号，把上面+改-

乘法不会，orz了某神犇之后才知道

f[i][j]+=f[i][k]*f[k+1][j]*C(j-i-1,k-i)

#include<cstdio>
#include<cstring>
#define LL long long
#define mod 1000000007
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int c[110][110];
LL a[110];
char s[110];
LL f[110][110];
LL jc[110];
inline void init()
{
    c[1][1]=1;
    for (int i=0;i<=100;i++)c[i][0]=1;
    for (int i=1;i<=100;i++)
        for (int j=1;j<=i;j++)
        c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
    jc[0]=1;
    for (int i=1;i<=100;i++)jc[i]=jc[i-1]*i%mod;
}
int n;
int main()
{
    init();
    while (~scanf("%d",&n))
    {
        memset(f,0,sizeof(f));
        for (int i=1;i<=n;i++)a[i]=read();
        scanf("%s",s+1);
        for (int i=1;i<=n;i++)f[i][i]=a[i];
        for (int len=1;len<=n;len++)
            for (int i=1;i<=n;i++)
            {
                int j=i+len-1;if (j>n)break;
                for (int k=i;k<j;k++)
                {
                    if (s[k]=='+')f[i][j]=(f[i][j]+(f[i][k]*jc[j-k-1]+f[k+1][j]*jc[k-i])%mod*c[j-i-1][k-i]%mod)%mod;
                    if (s[k]=='-')f[i][j]=(f[i][j]+(f[i][k]*jc[j-k-1]-f[k+1][j]*jc[k-i])%mod*c[j-i-1][k-i]%mod+mod)%mod;
                    if (s[k]=='*')f[i][j]=(f[i][j]+(f[i][k]*f[k+1][j])%mod*c[j-i-1][k-i])%mod;
                }
            }
        printf("%lld
",f[1][n]);
    }
}