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【AcWing】寒假每日一题

2021-04-29 18:59:06  阅读：135  来源： 互联网

标签：int res 每日 cin mid ++ 寒假 include AcWing


寒假每日一题

week 1

AcWing104.货仓选址

#include <algorithm>

using namespace std;

const int N = 100010;

int n, res;
int a[N];

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);

    sort(a, a + n);

    for (int i = 0; i < n; i ++ ) res += abs(a[i] - a[n / 2]);
    printf("%d\n", res);

    return 0;
}

week 2

AcWing756.蛇形矩阵

在这里插入图片描述

#include<iostream>

using namespace std;

const int N = 110;

int n, m;
int q[N][N];

int main()
{
    cin >> n >> m;
    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
    int x = 0, y = 0, d = 1;
    for (int i = 1; i <= n * m; i ++)
    {
        q[x][y] = i;
        int a = x + dx[d], b = y + dy[d];
        if (a < 0 || a >= n || b < 0 || b >= m || q[a][b])
        {
            d = (d + 1) % 4;
            a = x + dx[d], b = y + dy[d];
        }
        x = a, y = b;  
    }
    for (int i = 0; i < n; i ++)
    {
        for (int j = 0; j < m; j ++)
            cout << q[i][j] << ' ';
        cout << endl;
    }

    return 0;
}

AcWing1113.红与黑

BFS

#include<iostream>
#include<queue>
#include<algorithm>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;
const int N = 25;

int n, m;
char g[N][N];

int bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    g[sx][sy] = '#';
    int res = 0;

    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

    while (q.size())
    {
        auto t = q.front();
        q.pop();
        res ++ ;

        for (int i = 0; i < 4; i ++)
        {
            int x = t.x + dx[i], y = t.y + dy[i];
            if (x < 0 || x >= n || y < 0 || y >= m || g[x][y] != '.') continue;
            q.push({x, y});
            g[x][y] = '#';
        }
    }
    return res;
}

int main()
{
    while(cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i ++) cin >> g[i];
        int x, y;
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < m; j ++)
                if (g[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
        cout << bfs(x, y) << endl;
    }

    return 0;
}

DFS

#include<iostream>

using namespace std;

const int N = 25;

int n, m;
char g[N][N];

int dfs(int x, int y)
{
    int res = 1;
    g[x][y] = '#';
    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
    
    for (int i = 0; i < 4; i ++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m || g[a][b] != '.') continue;
        res += dfs(a, b);
    }
    return res;
}

int main()
{
    while(cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i ++) cin >> g[i];
        int x, y;
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < m; j ++)
                if (g[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
        cout << dfs(x, y) << endl;
    }

    return 0;
}

AcWing1346.回文平方

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

char get(int x)//将x转换为字符形式
{
    if (x <= 9) return x + '0';
    return x - 10 + 'A';
}

string base(int n, int b)//将n转换为b进制，返回对应字符串
{
    string res;
    while (n)
    {
        res += get(n % b);
        n /= b;
    }
    reverse(res.begin(), res.end());//翻转回高位在左
    return res;
}

bool check(string s)//检查s是否是回文
{
    for (int i = 0, j = s.size() - 1; i < j; i ++, j -- )
        if (s[i] != s[j])
            return false;
    return true;
}

int main()
{
    int b;
    cin >> b;

    for (int i = 1; i <= 300; i ++ )//暴力
    {
        string num = base(i * i, b);
        if (check(num))
            cout << base(i, b) << ' ' << num << endl;
    }

    return 0;
}

思考：如何将其他进制转化为十进制？秦九韶算法

#include<iostream>

using namespace std;

int uget(char c)
{
    if (c <= '9') return c - '0';//'0' 代表 字符0 ,对应ASCII码值为 0x30 (也就是十进制 48)
    return c + 10 - 'A';
}
int base10(string num, int b)
{
    int res = 0;
    for (auto c: num)
        res = res * b + uget(c);
    return res;
}

int main()
{
    string a;
    cin >> a;
    cout << base10(a, 2);
    return 0;
}

AcWing1227分巧克力

区分是向上取整还是向下取整关键看mid，如果mid属于左区间则向下取整，mid属于右区间则向上取整。
在这里插入图片描述

//二分模板
bool check(int x) {/* ... */} // 检查x是否满足某种性质

// 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用：
int bsearch_1(int l, int r)
{
    while (l < r)
    {
        int mid = l + r >> 1; // 下取整
        if (check(mid)) r = mid;    // check()判断mid是否满足性质
        else l = mid + 1;
    }
    return l;
}
// 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
int bsearch_2(int l, int r)
{
    while (l < r)
    {
        int mid = l + r + 1 >> 1;// 上取整
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    return l; // 最后这里其实不过返回l 或者 r 都是一样，因为只有相等的时候才会结束。
}

题目代码：

//cpp
#include<iostream>

using namespace std;

typedef long long LL;
const int N = 100010;

int n, m;
int h[N], w[N];

bool check(int mid)
{
    LL res = 0; //res 别忘了初始化
    for (int i = 0; i < n; i ++)
        res += (LL)h[i] / mid * (w[i] / mid);
   	return res >= m;
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> w[i] >> h[i];

    int l = 1, r = 1e5;
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }

    cout << r << endl;
    return 0;
}

AcWing680.剪绳子

二分的本质是将一个最优化问题转化为判定性问题。假定一个值是最优值，然后不断check 并更新此判定条件。每次只需判定下二分得到的值成不成立就可以

#include<iostream>

using namespace std;

const int N = 100010;

int n, m;
int w[N];

bool check(double mid)
{
    int cnt = 0;
    for (int i = 0; i < n; i ++)
        cnt += w[i] / mid;
    return cnt >= m;
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> w[i];

    double l = 0, r = 1e9;
    while (r - l > 1e-4)
    {
        double mid = (l + r) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }

    printf("%.2f\n", r);
    return 0;
}

AcWing429.奖学金

//解法一： 重载小于号
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 310;

int n;
struct Person
{
    int id, sum, a, b, c;
    bool operator< (const Person& t) const
    {
        if (sum != t.sum) return sum > t.sum;
        if (a != t.a) return a > t.a;
        return id < t.id;
    }
}q[N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        q[i] = {i, a + b + c, a, b, c};
    }

    sort(q + 1, q + n + 1);

    for (int i = 1; i <= 5; i++)
        cout << q[i].id << ' ' << q[i].sum << endl;
    
    return 0;

}

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010;

struct Student
{
    int id, a, b, c, sum;
    
    void calc()
    {
        sum = a + b + c;
    }
    
    bool operator< (const Student& t)const
    {
        if (sum != t.sum) return sum > t.sum;
        if (a != t.a) return a > t.a;
        return id < t.id;
    }
}q[N];

int main()
{
    int n;
    cin >> n;

    for (int i = 1; i <= n; i ++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        q[i].id = i;
        q[i].a = a;
        q[i].b = b;
        q[i].c = c;
        q[i].calc();
    }

    sort(q + 1, q + n + 1);

    for (int i = 1; i <= 5; i ++)
        cout << q[i].id << ' ' << q[i].sum << endl;
        
    return 0;
}

// 解法二
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 310;

int n;
struct Person
{
    int id, sum, a, b, c;
}q[N];

bool cmp(Person& a, Person& b)
{
    if (a.sum != b.sum) return a.sum > b.sum;
    if (a.a != b.a) return a.a > b.a;
    return a.id < b.id;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        q[i] = {i, a + b + c, a, b, c};
    }

    sort(q + 1, q + n + 1, cmp);

    for (int i = 1; i <= 5; i++)
        cout << q[i].id << ' ' << q[i].sum << endl;
    
    return 0;

}

week 3

AcWing422.校门外的树

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 10010;

int L, n;
bool st[N];

int main()
{
    cin >> L >> n;
    for (int i = 0; i <= L; i ++) st[i] = true;
    while (n --)
    {
        int a, b;
        cin >> a >> b;
        for (int i = a; i <= b; i ++) st[i] = false;
    }

    int res = 0;
    for (int i = 0; i <= L; i ++) res += st[i];
    cout << res << endl;

    return 0;
}

//区间合并算法
//将所有区间按左端点从小到大排序
//从左到右遍历每个区间

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 110;

int m, n;
struct Segment
{
    int l, r;
    bool operator< (const Segment& t) const
    {
        return l < t.l;
    }
}seg[N];

int main()
{
    cin >> m >> n;
    for (int i = 0; i < n; i ++) cin >> seg[i].l >> seg[i].r;
    sort(seg, seg + n);

    int sum = 0;
    int L = seg[0].l, R = seg[0].r;
    for (int i = 1; i < n; i ++)
        if (seg[i].l <= R) R = max(R, seg[i].r);
        else
        {
            sum += R - L + 1;
            L = seg[i].l, R = seg[i].r;
        }
    sum += R - L + 1;
    cout << m + 1 - sum << endl;

    return 0;
}

AcWing1381.阶乘

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main()
{
    int n;
    cin >> n;
    int res = 1, d2 = 0, d5 = 0;
    for (int i = 1; i <= n; i ++)
    {
        int x = i;
        while (x % 2 == 0) x /= 2, d2 ++;
        while (x % 5 == 0) x /= 5, d5 ++;
        res = res * x % 10; 
    }

    int k = min(d2, d5);
    for (int i = 0; i < d2 - k; i ++) res = res * 2 % 10;
    for (int i = 0; i < d5 - k; i ++) res = res * 5 % 10;
    cout << res << endl;

    return 0;
}

AcWing1208.翻硬币

#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

const int N = 310;

string a, b;

void turn(int i)
{
    if (a[i] == '*') a[i] = 'o';
    else a[i] = '*';
}

int main()
{
    cin >> a >> b;

    int res = 0;
    for (int i = 0; i + 1 < a.size(); i ++)
        if (a[i] != b[i])
        {
            res ++;
            turn(i), turn(i + 1);
        }
    cout << res << endl;
    return 0;
}

AcWing1532.找硬币

哈希表 哈希碰撞
在这里插入图片描述

#include<iostream>
#include<cstring>
#include<algorithm>
#include<unordered_set>

using namespace std;

const int INF = 100000;

int main()
{
    int n, m;
    cin >> n >> m;
    unordered_set<int> hash;

    int v1 = INF, v2;
    for (int i = 0; i < n; i ++)
    { 
        int a, b;
        cin >> a;
        b = m - a;
        if (hash.count(b))
        {
            hash.insert(a);
            if (a > b) swap(a, b);
            if (a < v1) v1 = a, v2 = b;
        }
        else hash.insert(a);
    }

    if (v1 == INF) puts("No Solution");
    else cout << v1 <<  ' ' << v2 << endl;

    return 0;
}
  • 双指针算法需要用到单调性
    判断能不能用双指针算法，只需看一个指针向后走的同时，另一个指针是不是一定往前走。
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 100010;

int n, m;
int w[N];

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> w[i];
    sort(w, w + n);

    for (int i = 0, j = n - 1; i < j; i ++)
    {
        while (i < j && w[i] + w[j] > m) j --;
        if (i < j && w[i] + w[j] == m)
        {
            cout << w[i] <<' '<< w[j] << endl;
            return 0;
        }
    }

    puts("No Solution");
    return 0;

}

在这里插入图片描述

AcWing1341.十三号星期五

  • 口诀：一三五七八十腊（12月），三十一天永不差；四六九冬（11月）三十日；平年二月二十八，闰年二月把一加。
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int weekday[7];

int main()
{
    int n;
    cin >> n;

    int days = 0;
    for (int year = 1900; year < 1900 + n; year ++)
    {
        for (int i = 1; i <= 12; i ++)
        {
            weekday[(days + 12) % 7] ++;
            days += month[i];
            if (i == 2)
            {
                if (year % 100 && year % 4 == 0 || year % 400 == 0)
                    days ++;
            }
        }
    }
//    for (int i = 5, j = 0; j < 7; i = (i + 1) % 7, j ++)
//        cout << weekday[i] << ' ';
	for (int j = 0, i = 5; j < 7; j ++)
   	{
        cout << weekday[i] << ' ';
        i = (i + 1) % 7;
    }
    cout << endl;
    
    return 0;
}

AcWing754.平方矩阵II

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 110;

int n;
int a[N][N];

int main()
{
    while (cin >> n, n)
    {
        for (int i = 1; i <= n; i ++)
        {
            for (int j = i, k = 1; j <= n; j ++, k ++)
            {
                a[i][j] = k;
                a[j][i] = k;
            }
        }

        for (int i = 1; i <= n; i ++)
        {
            for (int j = 1; j <= n; j ++)
                cout << a[i][j] << ' ';
            cout << endl;
        }
        cout << endl;
    }
    return 0;
}


#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 110;

int n;
int main()
{
    while (cin >> n, n)
    {
        for (int i = 1; i <= n; i ++)
        {
            for (int j = i; j >= 1; j --) cout << j << ' ';
            for (int j = i + 1; j <= n; j ++) cout << j - i + 1 << ' ';
            cout << endl;
        }
        cout << endl;
    }
}

AcWing1432.棋盘挑战

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 15;

int n;
bool col[N], dg[N * 2], udg[N * 2];
int path[N], ans;

void dfs(int x)
{
    if (x > n)
    {
        ans ++;
        if (ans <= 3)
        {
            for (int i = 1; i <= n; i ++)
                cout << path[i] << ' ';
            cout << endl;
        }
        return;
    }
    
    for (int y = 1; y <= n; y ++)
        if (!col[y] && !dg[x + y] && !udg[x - y + n])
        {
            path[x] = y;
            col[y] = dg[x + y] = udg[x - y + n] = true;
            dfs(x + 1);
            col[y] = dg[x + y] = udg[x - y + n] = false;
            path[x] = 0;
        } 
}

int main()
{
    cin >> n;
    dfs(1);
    cout << ans << endl;
    return 0;
}

AcWing1371.货币系统

DP问题

week 4

AcWing1353.滑雪场设计

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;

int n;
int h[N];

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> h[i];

    int res = 1e8;
    for (int i = 0; i + 17 <= 100; i ++)
    {
        int cost = 0, l = i, r = i + 17;
        for (int j = 0; j < n; j ++)
            if (h[j] < l) cost += (l - h[j]) * (l - h[j]);
            else if (h[j] > r) cost += (h[j] - r) * (h[j] - r);
        res = min(res, cost);
    }

    cout << res << endl;

    return 0;
}

AcWing1603.整数集合划分

AcWing482.合唱队形

动态规划

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int n;
int h[N];
int f[N], g[N];

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &h[i]);
    for (int i = 1; i <= n; i ++ )
    {
        f[i] = 1;
        for (int j = 1; j < i; j ++ )
            if (h[j] < h[i])
                f[i] = max(f[i], f[j] + 1);
    }

    for (int i = n; i; i -- )
    {
        g[i] = 1;
        for (int j = n; j > i; j -- )
            if (h[j] < h[i])
                g[i] = max(g[i], g[j] + 1);
    }

    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res, f[i] + g[i] - 1);

    printf("%d\n", n - res);

    return 0;
}

AcWing420.火星人

在这里插入图片描述

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 10010;

int n, m;
int w[N];

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++) cin >> w[i];
    while (m --)
    {
        int k = n;
        while (w[k - 1] > w[k]) k --;
        int t = k;
        while (w[t + 1] > w[k - 1]) t ++;
        swap(w[k - 1], w[t]);
        reverse(w + k, w + n + 1); 
    }
    for (int i = 1; i <= n; i ++) cout << w[i] << ' ';
    cout << endl;

    return 0;
}

标签：int,res,每日,cin,mid,++,寒假,include,AcWing
来源： https://blog.csdn.net/wei2019_/article/details/112709193

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