hdu 5396 区间DP
Expression
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 482 Accepted Submission(s): 284
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 .
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.
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
3 3 2 1 -+ 5 1 4 6 8 3 +*-*
Sample Output
2 999999689HintTwo numbers are considered different when they are in different positions.
Author
xudyh
Source
2015 Multi-University Training Contest 9
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define LL long long
const int mod = 1000000000 + 7;
const int N = 105;
LL d[N][N];
char s[N];
int n;
LL C[N][N];
LL A[N];
void doit()
{
A[0] = 1;
for(int i = 1; i <= 100; i++)
A[i] = (A[i - 1] * i) % mod;
memset(C, 0, sizeof C);
for(int i = 0; i <= 100; 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;
}
}
int main()
{
doit();
while(~scanf("%d", &n))
{
for(int i = 1; i <= n; i++) scanf("%I64d", &d[i][i]);
scanf("%s", s + 1);
for(int L = 1; L <= n; L++)
for(int i = 1; i + L <= n; i++)
{
int j = i + L;
d[i][j] = 0;
for(int k = i; k < j; k++)
{
LL tmp;
if(s[k] == '*') tmp = (d[i][k] * d[k + 1][j]) % mod;
else if(s[k] == '+') tmp = (d[i][k] * A[j - k - 1] + d[k + 1][j] * A[k - i]) % mod;
else tmp = (d[i][k] * A[j - k - 1] - d[k + 1][j] * A[k - i]) % mod;
d[i][j] = (d[i][j] + tmp * C[j - i - 1][k - i]) % mod;
}
}
printf("%I64d\n", (d[1][n] + mod) % mod);
}
return 0;
}