蓝桥杯百校真题大联赛第3期（四）

A组

外卖店的优先级：模拟

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

#define x first
#define y second

using namespace std;
typedef pair<int, int> PII;

const int N = 100010;

int n, m, t;
PII menu[N];
int last[N], score[N];
bool st[N];

int main()
{
    cin >> n >> m >> t;
    
    for (int i = 0; i < m; i ++ )
    {
        scanf("%d%d", &menu[i].x, &menu[i].y);
    }
    
    sort(menu, menu + m);
    
    for (int i = 0; i < m; i ++ )
    {
        int x = menu[i].x, y = menu[i].y;
        
        if (x - last[y] > 1)
        {
            score[y] -= x - last[y] - 1;
            if (score[y] < 0) score[y] = 0;
            if (score[y] <= 3) st[y] = false;
        }
        
        score[y] += 2;
        if (score[y] > 5) st[y] = true;
        last[y] = x;
    }
    
    int cnt = 0;
    
    for (int i = 1; i <= n; i ++ )
    {
        if (last[i] != t)
        {
            score[i] -= t - last[i];
            if (score[i] <= 3) st[i] = false;
        }
        
        if (st[i])
            cnt ++ ;
    }
    
    cout << cnt << endl;
    
    return 0;
}

修改数组：并查集

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

using namespace std;
const int N = 1100010;

int n;
int p[N];

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main()
{
    cin >> n;
    
    for (int i = 1; i < N; i ++ ) p[i] = i;
    
    for (int i = 0; i < n; i ++ )
    {
        int x;
        scanf("%d", &x);
        x = find(x);
        printf("%d ", x);
        p[x] = x + 1;
    }
    return 0;
}

B组

回文日期：模拟

错误思路：

这道题当时做的时候被坑了好久，原因是当时从下一年开始枚举的，其实要从当年开始枚举，因为每一年回文日期是一个确定的数，所以我们只需要枚举年份就行，然后判断是否回文。还有一个坑的地方就是最后的A和B不能相同，因此需要特判

#include <iostream>
#include <cstring>
#include <algorithm>
typedef long long LL;

using namespace std;
int res, res2;
int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool is_leap(int y, int m)
{
    if (m != 2) return false;
    return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;
}

bool check(int y, int m, int d)
{
    if (m < 1 || m > 12) return false;
    if (d < 1 || (d > months[m] + is_leap(y, m))) return false;
    return true;
}

int flip(int x)
{
    int res = 0;
    while (x)
    {
        res = res * 10 + x % 10;
        x /= 10;
    }
    return res;
}

int main()
{
    int n;
    cin >> n;
    int y = n / 10000;
    for (int i = y; i <= 9999; i ++ )
    {
        int yy = i, mm = 0, dd = 0;
        
        int t = flip(yy);
        dd = t % 100;
        mm = t / 100;
        
        t = yy * 10000 + mm * 100 + dd;
        if (t <= n) continue;
        if (!res && check(yy, mm, dd)) res = yy * 10000 + mm * 100 + dd;
        
        if (mm == dd && dd % 10 != dd / 10 && check(yy, mm, dd)) 
        {
            res2 = yy * 10000 + mm * 100 + dd;
            break;
        }
    }
    printf("%d\n%d\n", res, res2);
    return 0;
}

子串分值和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

const int N = 1e5 + 10;
char str[N];
typedef long long LL;
int pos[26];//记录每个字符最后一次出现的位置
int f[N];

int main() {
    scanf("%s", str + 1);
    int n = strlen(str + 1);
    LL ans = 0;
    for(int i = 1; i <= n; ++i) {
        int t = str[i] - 'a';
        f[i] = f[i - 1] + i - pos[t];
        pos[t] = i;
        ans += f[i];
    }

    printf("%lld\n", ans);
    return 0;
}

C组

最少砝码

// 从前i个砝码中选，重量为j的最少砝码的个数
// 数据范围较大，可能是个数学题
#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int k;
    cin >> k;
    cout << ceil(log(2*k+1)/log(3)) << endl;
}

杨辉三角形：二分+找规律

1. 每一次乘积的时候都要判断会不会越界
2. 找规律的题多写几行
3. 求组合数的代码
4. 判边界
5. 多测试几个数据

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

typedef long long LL;

using namespace std;

int n;

LL C(int a, int b)
{
    LL res = 1;
    for (int i = a, j = 1; j <= b; i --, j ++ )
    {
        res = res * i / j;
        if (res > n) return res;
    }
    return res;
}

bool check(int k)
{
    LL l = 2 * k, r = max(LL(n), l);
    while (l < r)
    {
        LL mid = l + r >> 1;
        if (C(mid, k) >= n) r = mid;
        else l = mid + 1;
    }
    if (C(r, k) != n) return false;
    
    cout << r * (r + 1) / 2 + k + 1 << endl;
    return true;
}

int main()
{
    cin >> n;
    for (int k = 16; ; k -- )
        if (check(k))
            break;
    return 0;
}