小米秋招筆試-算法(2019屆)

小米秋招筆試-算法(2019屆)

第一題

• 題目是輸入一個數組以及一個數，確定這個數是否能由這個數組中的某些數相加得到，輸出1或者0，表示可行或不可行。

• 思路：先排序，再用bfs。

• 代碼

    #include <cstdlib>
    #include <string>
    #include <iostream>
    #include <fstream>
    #include <sstream>
    #include <unordered_map>
    #include <unordered_set>
    #include <map>
    #include <set>
    #include <stdio.h>
    #include <numeric>
    #include <algorithm>
    #include <functional>
    #include <stack>
    #include <queue>
    #include <cmath>
    #include <assert.h>
    #include <vector>
    #include <memory>
    #include <memory.h>
    using namespace std;
    typedef long long ll;
    const int MOD = 1e9 + 7;
    typedef unsigned char uchar;
    
    // #define G_DEBUG
    // 定義unordered_set<pair<int,int>, pairhash> sets時會用到
    struct pairhash {
    public:
        template <typename T, typename U>
        std::size_t operator()(const std::pair<T, U> &x) const
        {
            return std::hash<T>()(x.first) ^ std::hash<U>()(x.second);
        }
    };
    
    bool bfs( vector<int>& nums, int idx, int curr, int goal )
    {
        if (curr == goal)
            return true;
        if (idx >= nums.size() || curr > goal)
            return false;
    
        if (nums[idx] + curr > goal)
            return false;
    
        bool ret = bfs( nums, idx+1, curr+nums[idx], goal );
        if (!ret)
            ret = bfs(nums, idx + 1, curr, goal);
        return ret;
    }
    
    int main()
    {
    
    #ifdef G_DEBUG
        // 調試使用
        int N = 0;
        ifstream file("data.txt");
        file >> N;
        vector<int> nums( N, 0 );
        for (int i = 0; i < N; i++)
            file >> nums[i];
        int goal = 0;
        file >> goal;
    #else
        vector<string> strs(15);
        int N = 0;
        cin >> N;
        vector<int> nums( N, 0 );
        for (int i = 0; i < N; i++)
            cin >> nums[i];
    
        int goal = 0;
        cin >> goal;
    #endif
        
        std::sort( nums.begin(), nums.end() );
        int ret = bfs( nums, 0, 0, goal );
    
        cout << ret << endl;
    
        system("pause");
        return 0;
    }
  

第二題

• leetcode上有原題，當時也沒想着搜索一下，沒做出來。。

• 題目鏈接：https://blog.csdn.net/liuchonge/article/details/78346951

• 思路：使用二分法求解。

• 代碼

    #include <cstdlib>
    #include <string>
    #include <iostream>
    #include <fstream>
    #include <sstream>
    #include <unordered_map>
    #include <unordered_set>
    #include <map>
    #include <set>
    #include <stdio.h>
    #include <numeric>
    #include <algorithm>
    #include <functional>
    #include <stack>
    #include <queue>
    #include <cmath>
    #include <assert.h>
    #include <vector>
    #include <memory>
    #include <memory.h>
    using namespace std;
    typedef long long ll;
    const int MOD = 1e9 + 7;
    typedef unsigned char uchar;
    
    #define G_DEBUG
    // 定義unordered_set<pair<int,int>, pairhash> sets時會用到
    struct pairhash {
    public:
    	template <typename T, typename U>
    	std::size_t operator()(const std::pair<T, U> &x) const
    	{
    		return std::hash<T>()(x.first) ^ std::hash<U>()(x.second);
    	}
    };
    
    
    bool canSplit( vector<int> &nums, int goal, int m )
    {
    	int count = 1;
    	ll total = 0;
    	for (int i = 0; i < nums.size(); i++)
    	{
    		total += nums[i];
    		if (total > goal)
    		{
    			total = nums[i];
    			count++;
    			if (count > m)
    				return false;
    		}
    	}
    
    	return true;
    }
    
    
    int main()
    {
    
    #ifdef G_DEBUG
    	// 調試使用
    	ifstream file("data.txt");
    	int N = 0, M = 0;
    	file >> N >> M;
    	vector<int> nums(N, 0);
    	for (int i = 0; i < N; i++)
    		file >> nums[i];
    #else
    	int N = 0, M = 0;
    	cin >> N >> M;
    	vector<int> nums( N, 0 );
    	for (int i = 0; i < N; i++)
    		cin >> nums[i];
    #endif
    
    	int left = 0, right = 0;
    	for (int i = 0; i < nums.size(); i++)
    	{
    		left = max(left, nums[i]);
    		right += nums[i];
    	}
    		
    
    	int mid = 0;
    	while (left < right)
    	{
    		mid = (left + right) / 2;
    		if (canSplit(nums, mid, M))
    			right = mid;
    		else
    			left = mid + 1;
    			
    	}
    
    	cout << left << endl;
    	
    
    
    	system("pause");
    	return 0;
    }