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2019 Haidian District Youth Programming Challenge Activity Elementary Group Rematch Test Questions Detailed Answers

2022-08-04 18:01:00 【51CTO】

Evaluation results of six programs：

2019Detailed answers to the primary school group rematch test questions of the Haidian District Youth Programming Challenge Activity_ios

1 约数

       
       #include <iostream>
       
#include <cmath>
       
using namespace std;
       
int main()
       
{
       
    int n;
       
    cin >> n;
       
    int root = sqrt(n);
       
    for(int i = 2; i <= root; i++)
       
    {
       
        if(0 == n % i)
       
        {
       
            cout << n / i;
       
            return 0;
       
        }
       
    }
       
    cout << 1;
       
    return 0;
       
}
      
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2 阶乘

       
       #include<iostream>
       
using namespace std;
       
// zero count from tail of n!
       
int zeroCnt1(int n)
       
{
       
    int cnt = 0;
       
    while(n)
       
    {
       
        n /= 5;
       
        cnt += n;
       
    }
       
    return cnt;
       
}
       
// zero count from tail of n!! when n is even
       
int zeroCnt2(int n)
       
{
       
    n /= 10;
       
    int cnt = n;
       
    while(n)
       
    {
       
        n /= 5;
       
        cnt += n;
       
    }
       
    return cnt;
       
}
       
int main()
       
{
       
    int n;
       
    cin >> n;
       
    cout << zeroCnt1(n) << ' ';
       
    if(n % 2)
       
    {
       
        cout << 0;
       
    }
       
    else
       
    {
       
        cout << zeroCnt2(n) << endl;
       
    }
       
    return 0;
       
}
      
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3 序列

       
       #include <iostream>
       
using namespace std;
       
int a[200005];
       
int main()
       
{
       
    int n, i;
       
    cin >> n;
       
    for(i = 0; i < n; i++)
       
    {
       
        cin >> a[i];
       
    }
       
    for(i = n - 1; i >= 0; i -= 2)
       
    {
       
        cout << a[i] << " ";
       
    }
       
    if(n % 2)
       
    {
       
        i = 1;
       
    }
       
    else
       
    {
       
        i = 0;
       
    }
       
    for(; i < n; i += 2)
       
    {
       
        cout << a[i] << " ";
       
    }
       
    return 0;
       
}
      
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4 糖果

       
       #include <iostream>
       
#include <memory.h>
       
#include <cstring> 
       
using namespace std;
       
char a[10000000 + 5];
       
inline string read()//inlineKeep going faster
       
{
       
    // getchar()Compare when the amount of data is largescanf和cin快
       
    char ch = getchar();
       
    string res = "";
       
    while(ch>='A' && ch<='Z')
       
    {
       
        res += ch;
       
        ch = getchar();
       
    }
       
    return res;
       
}
       
int main()
       
{
       
    freopen("candy.in", "r", stdin);
       
    freopen("candy.out", "w", stdout);
       
    int cnt[128];
       
    memset(cnt, 0, sizeof(cnt));
       
    int start = 0;
       
    int maxLen = 26;
       
    int len;
       
    bool same = false;
       
    //string a = read();
       
    scanf("%s", a);
       
    int Size = strlen(a);
       
    for(int i = 0; i < Size; i++)
       
    {
       
        cnt[a[i] - 65]++;
       
        if(2 == cnt[a[i] - 65])
       
        {
       
            same = true;
       
            for(int j = start; j < i; j++)
       
            {
       
                if(a[j] == a[i])
       
                {
       
                    len = i - j;
       
                    if(len < maxLen)
       
                    {
       
                        maxLen = len;
       
                    }
       
                    start = j + 1;
       
                    break;
       
                }
       
            }
       
            cnt[a[i] - 65] = 1;
       
        }
       
    }
       
    if(!same)
       
    {
       
        cout << "-1" << endl;
       
        return 0;
       
    }
       
    cout << maxLen << endl;
       
    return 0;
       
}
      
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5 迷宫

       
       #include <iostream>
       
using namespace std;
       
const int N = 205;
       
// 4 directions: right, left, down, up
       
const int dx[4] = {0, 0, 1, -1};
       
const int dy[4] = {1, -1, 0, 0};
       
int n, m, ans;
       
char a[N][N];
       
bool vis[N][N];
       
void dfs(int x, int y)
       
{
       
    for(int dir = 0; dir < 4; dir++)
       
    {
       
        int nextX = x + dx[dir];
       
        int nextY = y + dy[dir];
       
        if(!vis[nextX][nextY] && nextX >= 1 && nextX <= n && nextY >= 1 && nextY <= m && (a[nextX][nextY] == '.' || a[nextX][nextY] == '*'))
       
        {
       
            vis[nextX][nextY] = true;
       
            if(a[nextX][nextY] == '*')
       
            {
       
                ans++;
       
            }
       
            dfs(nextX, nextY);
       
        }
       
    }
       
}
       
int main()
       
{
       
freopen("maze.in", "r", stdin);
       
freopen(“maze.out”, “w”, stdout);
       
    int startX, startY;
       
    cin >> n >> m; // n rows m columns
       
    int row, col;
       
    for(row = 1; row <= n; row++)
       
    {
       
        for(col = 1; col <= m; col++)
       
        {
       
            cin >> a[row][col];
       
            if('S' == a[row][col])
       
            {
       
                startX = row;
       
                startY = col;
       
            }
       
        }
       
    }
       
    vis[startX][startY] = true;
       
    dfs(startX, startY);
       
    cout << ans << endl;
       
    return 0;
       
}
      
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6 盒子

       
       #include <iostream>
       
#include <cstdio>
       
#include <algorithm>
       
#include <set>
       
using namespace std;
       
const int N = 500000 + 5;
       
int n, a[N];
       
int main()
       
{
       
    freopen("box.in", "r", stdin);
       
    freopen("box.out", "w", stdout);
       
    scanf("%d", &n);
       
    for(int i = 1; i <= n; i++)
       
    {
       
        scanf("%d", &a[i]);
       
    }
       
    sort(a + 1, a + 1 + n);
       
    multiset<int, greater<int> > s; // Each element represents how many elements each heap has
       
    for(int i = 1; i <= n; i++)
       
    {
       
        // For descending collections,lower_boundWhat you get is the first less thana[i]的元素的位置
       
        // Only the number of heaps is less than or equal toa[i],a[i]to be added to the bottom of the heap
       
        // If the number of min heap is greater than a[i],Description to bea[i]新建一堆
       
        multiset<int>::iterator it = s.lower_bound(a[i]);
       
        if(it == s.end()) // sThere is no such element in it
       
        {
       
            // Add a new heap,堆的大小为1（contains this element）
       
            s.insert(1);
       
        }
       
        else
       
        {
       
            // Put elements into this heap,The number of elements in the heap is added1
       
            int cnt = *it + 1;
       
            s.erase(it);
       
            s.insert(cnt);
       
        }
       
    }
       
    printf("%d", s.size());
       
    return 0;
       
}
      
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