训练指南-数论-课后入门习题

最新推荐文章于 2022-09-28 16:02:40 发布

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本文集锦了多种算法技巧，包括最小公倍数、全加和、幂和阶乘等复杂问题的解决方法，通过示例代码详细解析了算法背后的逻辑与实现细节。

- 最小公约数和最大公倍数

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cmath>
using namespace std;

long long  G, L;
int main() {
	int t;
	cin >> t;
	while (t--) {
		cin >> G >> L;
		if (L%G == 0)
			cout << G << " " << L << '\n';
		else
			cout << -1 << '\n';
	}
	return 0;
}

–最小公倍数

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long ll;
int a, c;

int gcd(int a, int b){
	return b == 0 ? a : gcd(b, a%b);
}
int main() {
	int t;
	cin >> t;
	while (t--&&cin >> a >> c) {
		if (c%a)  cout << "NO SOLUTION\n";
		else {
			int b = c / a,r;
			while ((r=gcd(a, b)) != 1)
				a /= r, b = c / a;
			cout << b << '\n';
		}
	}
	return 0;
}

- 全加和
事先打表

#include<iostream>
#include<algorithm>
#include<string.h>
#define maxn 100
using namespace std;

long long d[maxn + 1][maxn + 1];
const int mod = 1000000;

int main() {
	memset(d, 0, sizeof(d));
	for (int i = 0; i <= maxn; i++) {
		d[i][1] = 1;
	}
	for (int k = 2; k <= maxn; k++) {
		for (int i = 0; i <= maxn; i++) {
			for (int n = 0; n <= maxn; n++)	  if (n - i >= 0) {
				d[n][k] =(d[n][k]+d[n - i][k - 1]) % mod;
			}
		}
	}
	int n, k;
	while (cin >> n >> k&&n&&k) {
		cout << d[n][k] << '\n';
	}
	return 0;
}

还有一种更快的方法，不过对于k个数的累加至n，我们用状态d(n,k)表示，
~~递推公式d(n,k)=sum{d(n,k-1)+d(n-1,k-1)} d(n,k-1) 表示与0搭配的顺序，d(n-1,k-1)表示与1搭配的顺序; 也可以考虑组合数的递推公式~~ 不解；

#include<iostream>
#include<algorithm>
#define N 205
using namespace std;

typedef long long ll;
const int MOD = 1000000;
int n, k;

ll C[N][N];

void init()
{
	memset(C, 0, sizeof(C));
	C[0][0] = 1;
	for (int i = 1; i < 200; i++)
	{
		C[i][0] = C[i][i] = 1;
		for (int j = 1; j < i; j++)
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
	}
}
int main()
{
	init();
	while (~scanf("%d%d", &n, &k))
	{
		if (!n && !k)break;
		int ans = C[n + k - 1][k - 1];
		printf("%d\n", ans);
	}
	return 0;
}

- 幂和阶乘

#include<iostream>
#include<algorithm>
#include<string.h>
#define maxn 10000
#define maxm 5000
using namespace std;

int cal(int n, int p) {
	int cur = 0;
	for (int i = 2; i <= n; i++) {
		int k = i;
		while (k%p == 0) {
			cur++;
			k = k / p;
		}
	}
	return cur;
}
int main() {

	int n, m,t,kase=1;
	cin >> t;
	while (t--&&cin >> m >> n){
		cout << "Case " << kase++ << ":\n";
		int ans = maxm;
		for (int j = 2; m > 1; j++) {
			int pnum = 0;
			while (m%j == 0) { pnum++, m /= j; }
			if (pnum) {
				int tt = cal(n, j) / pnum;
				if (ans > tt)  ans = tt;
			}
		}

		if (ans)
			cout << ans << '\n';
		else
			cout << "Impossible to divide\n";
	}
	return 0;
}

- LCM的个数

//int 型最大值为4294967296

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cmath>
#define maxn 10000+10
using namespace std;

typedef long long ll;
 int  num[maxn];

int gcd(int a, int b) {
	return b == 0 ? a : gcd(b, a%b);
}
int main() {
	int n;
	while (cin >> n && n) {
		cout << n << " ";
		if (n == 1) { cout << 1 << '\n'; continue; }
		int tot = 0;
		for (int i = 2; i < sqrt(n + 0.5); i++) {
			if (n%i == 0)
				num[tot++] = i, num[tot++] = n / i;
		}
		sort(num, num + tot);
		tot = unique(num, num + tot) - num;
		int coun = tot+2;   //预处理有与自身有关的搭配
		for (int i = 0; i < tot; i++) 
			for (int j = tot - 1; j >=i; j--)
				if ((((ll)num[i] * num[j]) / gcd(num[j], num[i])) == n) 
					coun++;
		cout << coun << '\n';
	}
	return 0;
}

- 超级幂

#include<iostream>
#include<algorithm>
#include<set>
#include<string.h>
#define maxn 64
using namespace std;

typedef unsigned long long ull;
const int p[18] = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61 };
ull lim = (1 << 64) - 1;
set<ull>s;

int main() {

	int vis[maxn];
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < 18; i++)  vis[p[i]] = 1;

	for (int i = 2;; i++) {
		ull   x = lim;
		int cnt = -1;
		while (x)   x /= i, cnt++;
		if (cnt < 4)  break;
		ull b = i;
		for (int j = 2; j <= cnt; j++) {
			b = b * i;
			if (!vis[j] && !s.count(b))   
				s.insert(b);
		}
	}
	s.insert(1);
	set<ull>::iterator it;   //迭代器访问;
	for (it = s.begin(); it != s.end(); it++)
		cout << *it << '\n';
	return 0;
}

- 排列之和
一个非常笨的方法 n>9以后就会非常慢


//运算量级太大;
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;

typedef unsigned long long ull;
set<ull> s;
int num[13],vis[13],help[13];
int n;
ull tran() {   //数组转化为数字;
	ull ans = 0;
	for (int i = 0; i < n; i++)
		ans = ans * 10 + help[i];
	return ans;
}

void dfs(int cur) {
	if (cur == n) {
		if (!s.count(tran()))  s.insert(tran());
	}
	else {
		for(int i=0;i<n;i++)
			if (!vis[i]) {
				help[cur] = num[i];
				vis[i] = 1;
				dfs(cur + 1);
				vis[i] = 0;
			}
	}
}

int main() {
	while (cin >> n && n) {
		s.clear();
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i <n; i++)  cin >> num[i];
		dfs(0);
		set<ull>::iterator it;
		ull ans = 0;
		for (it = s.begin(); it != s.end(); it++)
			ans += *it;
		cout <<ans<< '\n';
	}
	return 0;
}

另辟奇径网上搜的算法大佬
这个算法直到现在还是不理解啊，就这样，先记下来；

#include<iostream>
#include<algorithm>
#include <cmath>
using namespace std;

int n, x;
int vis[20];
long long fac[20];
long long one[13] = { 0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111,111111111, 1111111111, 11111111111, 111111111111 };
//这个数组用的很巧，佩服啊QAQ

int main() {
	fac[0] = 1;
	for (int i = 1; i <= 12; i++)
		fac[i] = fac[i - 1] * i;
	while (cin >> n && n) {
		memset(vis, 0, sizeof(vis));
		int sum = 0;
		long long  ans;
		for (int i = 0; i < n; i++) {
			cin >> x;
			vis[x]++;
			sum += x;
		}
		ans = fac[n - 1] * sum;
		for (int i = 0; i < 10; i++) {
			ans /= fac[vis[i]];
		}
		cout << ans * one[n] << '\n';
	}
	return 0;
}

- 组队
一时间没想起来

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;
const ll mod = 1000000007;
ll pow(int n, int p) {
	if (p == 1) { return n; }
	else{
	ll ans = pow(n, p / 2);
	ans = (ans*ans) % mod;
	if (p & 1) ans = (ans*n) % mod;
	return ans;}
}

int main() {
	int t, kase =1,n;
	cin >> t;
	while (t--) {
		cin >> n;
		cout << "Case #" << kase++ << ": ";
		if (n == 1)  cout << 1 << '\n';
		else    cout<< (n*pow(2, n-1)) % mod << '\n';
	}
	return 0;
}

- 回文数
转载大佬，也是不懂?


typedef long long ll;
ll p10[maxn], cnt[maxn], sum[maxn];
	
void inti() {    //初始化p10和sum数组;

	p10[0] = 1;
	for (int i = 1; i < maxn; i++) p10[i] = p10[i - 1] * 10;
	sum[0] = 0;
	for (int i = 1; i < maxn; i++) {
		cnt[i] = 9 * p10[(i - 1) / 2];
		sum[i] = sum[i - 1] + cnt[i];
	}
}

int ans[20], len;

void f(int i,int step) {   //递归数组;
	if (!step)  return;
	int j = len - 1 - i;
	int p = (j - i - 2) / 2+1;
	ans[i] = ans[j] = (step - 1) / p10[p];
	if (p)  f(i + 1, step - ans[i] * p10[p]);

}

int main() {

	inti();
	while(cin>>n&&n){
		len = 1;
		while (sum[len] < n)  len++;
		if (len == 1) { cout << n << '\n'; continue; }
		if (len == 2) { n -= 9; cout << n << n << '\n'; continue; }

		n -= sum[len-1];
		int p = (len - 3) / 2 + 1;
		ans[0] = ans[len - 1] = (n - 1) / p10[p] + 1;
		n -= (ans[0] - 1)*p10[p];
		if (n) f(1, n);
		for (int i = 0; i < len; i++)
			cout << ans[i];
		cout << '\n';
 	}
	return 0;
}

路漫漫其修远兮，吾将上下而求索