当前位置: 首页 > 工具软件 > Neko > 使用案例 >

Neko and Aki's Prank

袁何平
2023-12-01

https://codeforces.com/contest/1152/problem/D

题解:DP+规律 

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
ll dp[N][N],DP[N][N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    for(int i=1;i<=1000;i++){
        dp[i][0]=(i+1)/2;
        for(int j=1;j<=i;j++){
            dp[i][j]=(dp[i-1][j]+1+dp[i-1][j-1])%MOD;
            if(j>=2)dp[i][j]=(dp[i][j]+dp[i][j-2])%MOD;
            //printf("%9lld%c",dp[i][j]," \n"[j==i]);
        }
    }
    while(~scanf("%d",&n)){
        cout<<dp[n][n]<<endl;
    }

    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

 

 类似资料:

相关阅读

相关文章

相关问答