信息学奥赛一本通 2.4：递归算法

斐波那契数列

#include <iostream>
using namespace std;

int f(int a) {
    
    
	if (a==1 || a==2) return 1;
	return f(a-2) + f(a-1);
}
	
int main(){
    
    
	int n, a;
	cin >> n;
	
	while (n--) {
    
    
		cin >> a;
		cout << f(a) << endl;
	}

	return 0;
}

爬楼梯

#include <iostream>
using namespace std;

int f(int n) {
    
    
	if (n==1 || n==2) return n;
	return f(n-1) + f(n-2);
}

int main() {
    
    
	int n;
	
	while (cin >> n) {
    
    
		cout << f(n) << endl;
	}
	
	return 0;
}

Pell数列

#include <iostream>
using namespace std;

const int N=1e6+5, MOD=32767;
int a[N];

int f(int x) {
    
    
	if (a[x]) return a[x];

	if (x == 1) a[x] = 1;
	else if (x == 2) a[x] = 2;
	else a[x] = (f(x-1)*2 + f(x-2)) % MOD;

	return a[x];
}

int main() {
    
    
	int n;
	cin >> n;
	
	while (n --) {
    
    
		int k;
		cin >> k;
		cout << f(k) << endl;
	}

	return 0;
}

数的计数

#include <iostream>
using namespace std;

const int N=1005;
int a[N];

int f(int x) {
    
    
	if (a[x]) return a[x];

	if (x==0 || x==1) a[x] = 1;
	else a[x] = f(x-2) + f(x/2);

	return a[x];
}

int main() {
    
    
	int n;
	cin >> n;
	
	cout << f(n);

	return 0;
}

放苹果

#include <iostream>
using namespace std;

int f(int m, int n) {
    
    
	if (m==0 || n==1) return 1;
	if (m < n) return f(m, m);
	
	return f(m, n-1) + f(m-n, n);
}

int main() {
    
    
	int t, m, n;
	cin >> t;
	
	while (t--) {
    
    
		cin >> m >> n;
		cout << f(m, n) << endl;
	}
	
	return 0;
}

集合的划分

#include <iostream>
using namespace std;

long long s(int n, int k) {
    
    
	if(n==k || k==1) return 1;
	return k*s(n-1,k) + s(n-1,k-1);
}


int main() {
    
    
	int n, k;
	cin >> n >> k;
	
	cout << s(n,k) << endl;

	return 0;
}

判断元素是否存在

#include <iostream>
#include <cstdio>
using namespace std;

int x;

bool f(int k){
    
    
	if(k > x) return 0;
	if(k == x) return 1;
	
	return f(2*k+1) || f(3*k+1);
}

int main() {
    
    
	int k;
	scanf("%d,%d",&k, &x);
	
	if(f(k)) cout << "YES";
	else cout << "NO";

	return 0;
}

求最大公约数问题

#include <iostream>
using namespace std;

int gcd (int a, int b) {
    
    
	if (b == 0) return a;
	return gcd(b, a%b);
}

int main() {
    
    
	int a, b;
	cin >> a >> b;
	cout << gcd(a, b) << endl;
	
	return 0;
}

分数求和

#include <iostream>
#include <cstdio>
using namespace std;

int gcd(int x, int y) {
    
    
	if (y == 0) return x;
	
	return gcd(y, x%y);
}

int main() {
    
    
	int n;
	cin >> n;
	
	int a, b, c, d;

	scanf("%d/%d", &a, &b);
	for (int i = 2; i <= n; i ++) {
    
    
		scanf("%d/%d", &c, &d);
		
		int fz = a*d + b*c;
		int fm = b*d;
		
		int t = gcd(fz, fm);
		
		a = fz / t;
		b = fm / t;
	}
	 
	printf("%d/%d", a, b);	
	
	return 0;
}

全排列

#include <iostream>
using namespace std;

string s;
char ans[6];
bool used[6];
int len;

void dfs(int x) {
    
    
	if (x == len) {
    
    
		for (int i = 0; i < len; i ++) cout << ans[i];
		cout << endl;
		return;
	}
	
	for (int i = 0; i < len; i ++) {
    
    
		if (!used[i]) {
    
    
			ans[x] = s[i];
			used[i] = true;
			
			dfs(x + 1);
			
			used[i] = false;
		}
	}	
}


int main() {
    
    
	cin >> s;
	len = s.size();
	
	dfs(0);
	
	return 0;
}

因子分解

#include <iostream>
using namespace std;

int ans[40], len;

void f(int x) {
    
    
	if (x == 1) {
    
    
		cout << ans[1];
		
		int cnt = 1;
		for (int i = 2; i <= len; i ++) {
    
    
			if (ans[i] == ans[i-1]) cnt ++;
			else {
    
    
				if (cnt == 1) cout << '*' << ans[i];
				else cout << '^' << cnt << '*' << ans[i];
				cnt = 1;
			}
		}
		if (ans[len] == ans[len-1]) cout << '^' << cnt;
	}
	
	for (int i = 2; i <= x; i ++) {
    
    
		if (x%i == 0) {
    
    
			ans[++len] = i;
			f(x/i);
			break;
		}
	}
}


int main() {
    
    
	int n;
	cin >> n;
	
	f(n);
	
	return 0;
}

分解因数

#include <iostream>
using namespace std;

int a, cnt;
int ans[20] = {
    
    2};

void dfs(int x) {
    
    
	if (a == 1) {
    
    
		cnt ++;
		return;
	}
	
	for (int i = ans[x-1]; i <= a; i ++) {
    
    
		if (a % i == 0) {
    
    
			ans[x] = i;
			a /= i;
			
			dfs(x+1);
			
			a *= i;
		}
	}	
}

int main() {
    
    
	int n;
	cin >> n;
	
	while (n --) {
    
    
		cin >> a;
		
		cnt = 0;
		dfs(1);
		cout << cnt << endl;		
	}
	
	return 0;
}

2的幂次方表示

#include <iostream>
using namespace std;

int a[15] = {
    
    1};

void f(int x) {
    
    
	for (int i = 14; i >= 0; i --) {
    
    
		if (x >= a[i]) {
    
    
			if (i == 0) cout << "2(0)";
			else if (i == 1) cout << "2";
			else {
    
    
				cout << "2(";
				f(i);
				cout << ")";
			}
			
			x -= a[i];
			if (x > 0) cout << "+";
			else return;
		}
	}
}

int main() {
    
    
	int n;
	cin >> n;
	
	for (int i = 1; i <= 14; i ++) a[i] = a[i-1] * 2;
	
	f(n);

	return 0;
}

逆波兰表达式

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

char s[20];

double f() {
    
    
    scanf("%s", s);
    if (s[0] == '+') return f() + f();
	if (s[0] == '-') return f() - f();
	if (s[0] == '*') return f() * f();
	if (s[0] == '/') return f() / f();

	return atof(s);
}

int main() {
    
    
    printf("%f\n", f());
    
    return 0;
}

汉诺塔问题

#include <iostream>
#include <cstdio>
using namespace std;

int n;
char a,b,c;

void mov(int n, char a, char b, char c) {
    
    
	if (n == 0) return;
	
	mov(n-1, a, c, b);
	printf("%c->%d->%c\n", a, n, c);
	mov(n-1, b, a, c);
}

int main() {
    
    
	cin >> n >> a >> b >> c;
	
	mov(n, a, c, b);
	
	return 0;
}

扩号匹配问题

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

char a[100],b[100];

int up (int d) {
    
    
	if (d==-1) return -1;
	else if (b[d]=='$') return d;
	else return up(d-1);
}

int main() {
    
    
	while (cin>>a) {
    
    	
		cout << a << endl;
		
		int len = strlen(a);
		memset(b,' ',sizeof(b));
		
		for (int i=0;i<len;i++) {
    
    
			if (a[i]=='(') b[i]='$';
			else if (a[i]==')') {
    
    
				int m=up(i-1);
				if (m==-1) b[i]='?';
				else b[m]=' ';
			}
		}

		cout << b << endl;
	}
	
	return 0;
}