【蓝桥杯专题】递归 &递推（C++ | 洛谷 | acwing）

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【蓝桥杯真题】（一）递归与递推 (Acwing 蓝桥杯笔记)——链接

P5534 【XR-3】等差数列

等差数列公式
公式一：Sn = (a1 + an) × n / 2;
公式二：Sn = n × a1 + n × (n - 1) × d / 2;

#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include <stdlib.h> // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e6 + 10;


int a[N];

void solve () {
	ll n, d, ans = 0;
	cin >> a[0] >> a[1] >> n;
	d = a[1] - a[0];

	ans = n * a[0] + n * (n - 1) * d / 2;
	cout << ans << endl;
}

int main(void){
	freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int T = 1;
    // cin >> T;
	while(T --) solve();
	return 0;
}

P4994 终于结束的起点

lj
+

#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include <stdlib.h> // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e6 + 10;



ll M;
ll fp[10000002];


// int a[N];
ll f(ll i) {
	if(fp[i]) return fp[i];
	if(i == 1 || i == 2) return fp[i] = 1 % M;
	else return fp[i] = (f(i - 1) + f(i - 2)) % M;
}

void solve () {
	cin >> M;
	ll z  = 1;
	// rep(i, 2, 1000) {
		while(f(z)  !=0 || f(z + 1) != 1) {
			z ++;
		}
	cout << z << endl;

}

int main(void){
	freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int T = 1;
    // cin >> T;
	while(T --) solve();
	return 0;
}

P1028 [NOIP2001 普及组] 数的计算

链接链接
在这里插入图片描述


#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include <stdlib.h> // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e6 + 10;

int dp[1001];

int f(int n) {
  if(n == 1)return 1;
  else if(n == 2) return 2;
  if(n % 2 == 0) return f(n - 1) + f(n / 2);
  else if(n % 2 == 1) return f(n - 1);
}

void solve () {
    int n;
    cin >> n;
    // cout << f(n) << endl; //TLE

    rep(i, 1, n) {
        rep(j, 1, i / 2) {
            dp[i] += dp[j];
        }
        dp[i] ++;
    }
    cout << dp[n] << endl;

}

int main(void){
	freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int T = 1;
    // cin >> T;
	while(T --) solve();
	return 0;
}

波动数列

链接链接

#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include <stdlib.h> // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=100000007;
const int N = 1005;

int n , s, a, b;
int f[N][N];

// 那
int get_mod (int a, int b) {
    return (a % b + b) % b;
}

void solve () {
    cin >> n >> s >> a >> b;
    f[0][0] = 1;
      for(int i = 1 ; i < n ; i++)
        for(int j = 0 ; j < n ; j++)
            f[i][j] = (f[i - 1][get_mod(j - i * a , n)] + f[i - 1][get_mod(j + i * b, n)]) % mod; 
    cout << f[n - 1][get_mod(s, n)] << endl;
}

int main(void){
	freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int T = 1;
    // cin >> T;
	while(T --) solve();
	return 0;
}

[递归]母牛的故事

链接链接
在这里插入图片描述

#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include <stdlib.h> // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;

int f[56] = {0, 1, 2, 3, 4};


// ! 递归可能超时   建议 记忆化搜索
// 递归题  写前几项规律 或许可以碰运气
int dfs(int n) {
    if(n <= 4) return n;
    else if (f[n] == 0) {
        f[n] = dfs(n - 1) + dfs(n - 3); ///数组记录递归结果，加快速度,记忆递归

        return f[n];
    }
    else if (f[n] != 0) {
        return f[n];
    }

}

void solve () {
    int n;
    // cin >> n
    while(cin >> n) {
        // cin >> n;
        if(n == 0) break;
        // cout << n << endl;
        cout << dfs(n) << endl;
    }
}

int main(void){
	// freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int T = 1;
    // cin >> T;
	while(T --) solve();
	return 0;
}