Keep2019届校招算法笔试题

1. 输入一个字符串和整数k,每3k个子串，将子串前面的k个字符反转，后面2*k个子串保持原样。如果剩余的子串长度不足3k,则如果长度小于k,此末尾子串全部反转，如果剩余的子串长度大于k且小于3k,则把前面的k个子串翻转，后面的末尾字符都保持原样。

代码如下：

#include
#include
#include
using namespace std;

int main()
{
	string str;
	int k;
	cin >> str >> k;
	if (k <= 0) {
		cout << str << endl;
		return 0;
	}
	int n = str.size() / (3 * k);
	int t = str.size() - 3 * k*n;
	int i = 0, j = 0;
	for (; j < n; j++, i += 3 * k){
		reverse(str.begin() + i, str.begin() + i + k);
	}
	if (t0){
		reverse(str.begin() + i, str.end());
	}
	if (t >= k) {
		reverse(str.begin() + i,str.begin()+i+k);
	}
	cout << str << endl;
	return 0;
}

2.背包问题的变形，第一行输入容量Cap,和物体个数n，接下来n行表示n组物体体重和物体价值，求使得在此容量一定的条件下物体价值最大化的物品取法。1表示取物品，0表示不取物品。典型的动态规划问题，就是要加一个回溯函数。代码如下：

#include
#include
using namespace std;

int fun(int** f, int i,int cap, int n, int* weight, int* profit)
{
	if (f[i][cap] > 0) {
		return f[i][cap];
	}
	if (i == n) {
		f[i][cap] = weight[n] <= cap ? profit[i] : 0;
		return f[i][cap];
	}
	if (weight[i] > cap) {
		f[i][cap] = fun(f, i + 1, cap, n, weight, profit);
	}
	else {
		f[i][cap] = max(fun(f, i + 1, cap, n, weight, profit), fun(f, i + 1, cap - weight[i], n, weight, profit) + profit[i]);
	}
	return f[i][cap];
}

void traceback(int** f, int* weight, int cap, int n, int* x)
{
	for (int i = 1; i > cap>> n;
	int* profit = new int[n+1];
	int* weight = new int[n+1];
	for (int i =1; i <=n; i++) {
		cin >> weight[i] >> profit[i];
	}
	int** f = new int*[n+1];
	for (int i =0; i <=n; i++) {
		f[i] = new int[cap+1];
		for (int j = 0; j <= cap; j++) {
			f[i][j] = -1;
		}
	}
	int* x = new int[n + 1];
	//fun(f,cap,n,weight,profit);
	fun(f,1,cap,n,weight,profit);
	traceback(f,weight,cap,n,x);

	for (int i = 1; i <= n; i++) {
		cout << x[i] << endl;
	}

	delete [] profit;
	delete [] weight;
	for (int i = 0; i <=n; i++) {
		delete[] f[i];
	}
	delete [] f;
	return 0;
}

方法二，可以用迭代法写fun函数，代码如下

void fun(int** f, int cap, int n,int* weight,int* profit)
{
	int ymax = min(weight[n]-1, cap);
	for (int y = 0; y <=ymax; y++) {
		f[n][y] = 0;
	}
	for (int y =weight[n]; y <= cap; y++) {
		f[n][y] = profit[n];
	}
	for (int i = n-1; i>=1; i--) {
		ymax = min(weight[i]-1, cap);
		for (int y = 0; y <= ymax; y++) {
			f[i][y] = f[i + 1][y];
		}
		for (int y = weight[i]; y <= cap; y++) {
			f[i][y] = max(f[i + 1][y], f[i + 1][y - weight[i]] + profit[i]);
		}
	}
}

3.给定一个数字字符串，例如“1234”，输入一个整数k，例如4。求出所有可以被k整除的子串组合，此例输出12，124，24，4

（假设，子串的字符相对顺序是不能变化的，记不清题目了，，，，，如果是不管顺序，那就复杂了，先对字符串排序，然后调用bool next_permutation(iterator start,iterator end)）产生此字符串的下一排列，即对所有可能的排列进行递归。）

#include
#include
#include
#include
#include
using namespace std;

void dfs(string str,int b,string s,int k,set& ret,vector& used)
{
	if (!s.empty()&&stoi(s)%k == 0) {
		ret.insert(stoi(s));
	}
	for (int i = b; i < str.size(); i++) {
		if (used[i] == 0) {
			used[i] = 1;
			dfs(str, b + 1, s.append(1, str[b]), k, ret,used);
			s.pop_back();
			dfs(str, b + 1, s, k, ret,used);
			used[i] = 0;

		}
	}
}

int main()
{
	string str;
	int k;
	getline(cin, str);
	cin >> k;
	set ret;
	vector used(str.size(), 0);
	dfs(str,0,"",k,ret,used);
	for (int a : ret) {
		cout << a << endl;
	}
	system("pause");
	return 0;
}