目录

821. 跳台阶 - AcWing题库

​92. 递归实现指数型枚举 - AcWing题库 

​P1706 全排列问题 - 洛谷 (luogu.com.cn)

P1157 组合的输出 - 洛谷 (luogu.com.cn)

​P1036 [NOIP 2002 普及组] 选数 - 洛谷 (luogu.com.cn)

P2089 烤鸡 - 洛谷 (luogu.com.cn)

P1088 [NOIP 2004 普及组] 火星人 - 洛谷 (luogu.com.cn)

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 821. 跳台阶 - AcWing题库

#include<iostream>
using namespace std;
int t(int n) {
	if (n <= 2)return n;
	if (n >= 3)return t(n - 1) + t(n - 2);
}
int main() {
	int n; cin >> n;
	cout << t(n);
	return 0;
}

92. 递归实现指数型枚举 - AcWing题库 

#include<iostream>
using namespace std;
int n;
int s[20];//0表示未考虑;1表示选;2表示不选
void dfs(int x) {
	if (x > n) {
		for (int i = 1; i <= n; i++) {
			if (s[i] == 1)cout << i << " ";
		}
		cout << endl;
		return;
	}
	for (int i = 1; i <= 2; i++) {
		s[x] = i;
		dfs(x + 1);
	}
}
int main() {
	cin >> n;
	dfs(1);
	return 0;
}

 P1706 全排列问题 - 洛谷 (luogu.com.cn)

#include<iostream>
using namespace std;
int n;
int s[20];
bool vis[20];
void dfs(int x) {
	if (x > n) {
		for (int i = 1; i <= n; i++) {
			printf("%5d", s[i]);
		}
		cout << endl;
		return;
	}
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {
			vis[i] = true;
			s[x] = i;
			dfs(x + 1);
			vis[i] = false;
			s[x] = 0;
		}
	}
}
int main() {
	cin >> n;
	dfs(1);
	return 0;
}

 P1157 组合的输出 - 洛谷 (luogu.com.cn)

#include<iostream>
using namespace std;
int n, r;
int a[25];
void dfs(int x, int start) {
	if (x > r) {
		for (int i = 1; i <= r; i++) {
			printf("%3d", a[i]);
		}
		cout << endl;
		return;
	}
	for (int i = start; i <= n; i++) {
		a[x] = i;
		dfs(x + 1, i + 1);
		a[x] = 0;
	}
}
int main() {
	cin >> n >> r;
	dfs(1,1);
	return 0;
}

 P1036 [NOIP 2002 普及组] 选数 - 洛谷 (luogu.com.cn)

#include<iostream>
using namespace std;
int n, k;
int a[25]; int s[25];
int res;
bool jud(int t) {
	if (t < 2)return false;
	for (int i = 2; i * i <= t; i++) {
		if (t % i == 0)return false;
	}
	return true;
}
void dfs(int x, int start) {
	if (x > k) {
		int sum = 0;
		for (int i = 1; i <= k; i++) {
			sum += s[i];
		}
		if (jud(sum))res++;
		return;
	}
	for (int i = start; i <= n; i++) {
		s[x] = a[i];
		dfs(x + 1, i + 1);
		s[x] = 0;
	}
}
int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++)cin >> a[i];
	dfs(1, 1);
	cout << res;
	return 0;
}

P2089 烤鸡 - 洛谷 (luogu.com.cn)

#include<iostream>
using namespace std;
int n;
int res;
int a[60000][11];
int tem[11];
void dfs(int x, int sum) {
	if (sum > n)return;
	if (x > 10) {
		if (sum == n) {
			res++;
			for (int i = 1; i <= 10; i++) {
				a[res][i] = tem[i];
			}
		}
		return;
	}
	for (int i = 1; i <= 3; i++) {
		tem[x] = i;
		dfs(x + 1, sum + i);
	}
}
int main() {
	cin >> n;
	dfs(1, 0);
	cout << res << endl;
	for (int i = 1; i <= res; i++) {
		for (int j = 1; j <= 10; j++) {
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
	return 0;
}

 P1088 [NOIP 2004 普及组] 火星人 - 洛谷 (luogu.com.cn)

#include<iostream>
using namespace std;
int n, m;
int res;
const int N = 10010;
int a[N]; bool vis[N]; int mars[N];
void dfs(int x) {
	if (x > n) {
		res++;
		if (res == m + 1) {
			for (int i = 1; i <= n; i++) {
				cout << a[i] << " ";
			}
			exit(0);//剪枝，输出之后强制退出
		}
	}
	for (int i = 1; i <= n; i++) {
		if (!res) {
			i = mars[x];
		}
		if (!vis[i]) {
			vis[i] = true;
			a[x] = i;
			dfs(x + 1);
			vis[i] = false;
			a[x] = 0;
		}
	}
}
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)cin >> mars[i];
	dfs(1);
	return 0;
}

P1149 [NOIP 2008 提高组] 火柴棒等式 - 洛谷 (luogu.com.cn)

#include <iostream>
using namespace std;
int n;
int res;
const int N = 10010;
int a[N];
int num[N] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
int col(int p)
{
    if (num[p])
    {
        return num[p];
    }
    else
    {
        int t = 0;
        while (p)
        {
            t += num[p % 10];
            p /= 10;
        }
        return t;
    }
}
void dfs(int x, int sum)
{
    if (sum > n)
    {
        return;
    }
    if (x > 3)
    {
        if (a[1] + a[2] == a[3] && sum == n)
        {
            res++;
        }
        return;
    }
    for (int i = 0; i < N / 10; i++)
    {
        a[x] = i;
        dfs(x + 1, sum + col(i));
        a[x] = 0;
    }
}
int main()
{
    cin >> n;
    n -= 4;
    dfs(1, 0);
    cout << res;
    return 0;
}

P2036 [COCI 2008/2009 #2] PERKET - 洛谷 (luogu.com.cn) 

#include <iostream>
#include <cmath>
#include <climits>
using namespace std;
int n;
int s[11];
int b[11];
int t[11]; // 0:未考虑；1：选；2：不选
int res = INT_MAX;
bool flag;
void dfs(int x)
{
    if (x > n)
    {
        int x = 1;
        int y = 0;
        for (int i = 1; i <= n; i++)
        {
            if (t[i] == 1)
            {
                flag = true;
                x *= s[i];
                y += b[i];
            }
        }
        if (flag)
        {
            res = min(abs(x - y), res);
            flag = false;
        }
        return;
    }
    for (int i = 1; i <= 2; i++)
    {
        t[x] = i;
        dfs(x + 1);
        t[x] = 0;
    }
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i] >> b[i];
    }
    dfs(1);
    cout << res;
    return 0;
}