cf 495人题简单dp

最新推荐文章于 2023-11-26 20:16:46 发布

m0_37802215

最新推荐文章于 2023-11-26 20:16:46 发布

阅读量169

点赞数

CC 4.0 BY-SA版权

分类专栏： acm

本文链接：https://blog.youkuaiyun.com/m0_37802215/article/details/79102045

acm 专栏收录该内容

87 篇文章

订阅专栏

本文介绍了一种通过解析Python代码中的for循环结构来计算不同代码形式数量的方法。使用动态规划解决给定字符串序列对应的代码结构变化问题。

就是python的语法,for之后一定要缩进,给出for或者statement的顺序,问你代码形式有多少种.

/* Farewell. */
#include <iostream>
#include <vector>
#include <cstdio>
#include <stack>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <set>
using namespace std;
#define lson lmrt<<1
#define rson m+1rrt<<1|1
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define gcd __gcd
#define debug(x) std::cerr << #x << " = " << (x) << std::endl
typedef long long  LL;
typedef unsigned long long ULL;
typedef pair<int,int > pii;
typedef pair<LL,LL> pll;
typedef pair<double,double > pdd;
typedef pair<double,int > pdi;
const int INF = 0x7fffffff;
const LL INFF = 0x7f7f7f7fffffffff;
const int MAXM = 5e3+17;


const int MOD = 1e9+7;
const int MAXN = 5e3+17;
int dp[MAXN][MAXN];
char cmd[MAXN];
int main(int argc ,char const *argv[])
{
    #ifdef GoodbyeMonkeyKing
    freopen("Input.txt","r",stdin);freopen("Output.txt","w",stdout);
    #endif
    int n;
    cin>>n;
    for (int i = 0; i < n; ++i)
        cin>>cmd[i];
    dp[0][0] = 1;
    for (int i = 1; i < n; ++i)
    {
        if(cmd[i-1]=='f')
            for (int j = 0; j < n; ++j)
                dp[i][j+1] = dp[i-1][j];
        else 
        {
            int sum = 0;
            for (int j = n-1; j > -1; --j)
            {
                dp[i][j] = (sum+dp[i-1][j])%MOD;
                sum=(dp[i-1][j]+sum)%MOD;
            }
        }
    }
    int ans = 0;
    for (int i = 0; i <= n; ++i)
        ans=(dp[n-1][i]+ans)%MOD;
    cout<<ans<<endl;
    return 0;
}