2014 Super training #10 d peanut sequence-DP

Original question: fzu 2170 http://acm.fzu.edu.cn/problem.php? PID = 1, 2170

I did not understand the question at the time, and I didn't even think about the example. So I still feel that it is not easy to do this. Some questions are obviously very simple, but I gave up and even went to play because I didn't understand it. This would not achieve much effect.

Solution:

Definition: DP [I] [J]: The number of labeling schemes in which J of the first I letter belongs to the first sequence.

In case of 'B', the first sequence must be an odd number, that is, J & 1 is transferred. When the number of the second sequence is an odd number (I-j) & 1), it also needs to be transferred because B may belong to the second sequence. Otherwise.

Use a scrolling array to save space.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#define Mod 1000000007using namespace std;#define N 6007int dp[2][N];char ss[N];int main(){    int n,i,j;    int now;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        scanf("%s",ss);        memset(dp,0,sizeof(dp));        dp[0][0] = 1;        for(i=0,now=1;i<2*n;i++,now=1-now)        {            memset(dp[now],0,sizeof(dp[now]));            if(ss[i] == ‘B‘)            {                for(j=0;j<=n;j++)                {                    if(j&1)                        dp[now][j+1] = (dp[now][j+1]+dp[i&1][j])%Mod;                    if((i-j)&1)                        dp[now][j] = (dp[now][j]+dp[i&1][j])%Mod;                }            }            else            {                for(j=0;j<=n;j++)                {                    if((j&1) == 0)                        dp[now][j+1] = (dp[now][j+1]+dp[i&1][j])%Mod;                    if(((i-j)&1) == 0)                        dp[now][j] = (dp[now][j]+dp[i&1][j])%Mod;                }            }        }        printf("%d\n",dp[0][n]);    }    return 0;}

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