AtCoder Beginner Contest 127 解题报告

 

传送门

非常遗憾。当天晚上错过这一场。不过感觉也会掉分的吧。后面两题偏结论题，打了的话应该想不出来。

 

A - Ferris Wheel

#include 
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int main() {
    int a = read(), b = read();
    if (a >= 13) b = b;
    else if (a > 5) b /= 2;
    else b = 0;
    cout << b << '\n';
    return 0; 
}

View Code

 

B - Algae

#include 
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int main() {
    int r = read(), d = read(), x = read();
    for (int i = 0; i < 10; i++) {
        x = r * x - d;
        printf("%d\n", x);
    }
    return 0;
}

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C - Prison

题意：有$N$张ID卡，编号为1到$N$，$M$扇门，每扇门对应着一个区间$\left[ L,R\right]$，在闭区间内的ID卡才能打开门，问有多少张ID卡能打开所有门。

思路：相当于$M$个区间求交集。然后就是求$L$的最大值和$R$的最小值，$L$大于$R$就无解，否则就是$R - L + 1$

#include 
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int main() {
    int n = read(), m = read();
    int l = 1, r = n;
    while (m--) {
        int u = read(), v = read();
        l = max(l, u); r = min(r, v);
    }
    int ans;
    if (l > r) ans = 0;
    else ans = r - l + 1;
    printf("%d\n", ans);
    return 0;
}

View Code

 

D - Integer Cards

题意：给$N$张卡片，有$M$次操作，每次可以至多把$B$张卡片上的数改成$C$，问最后$N$个数的和最大是多少。

思路：可以证(举例)明(发现)，最后结果与操作顺序无关。

那么就把$N$个数放进小根堆，然后把操作按$C$ 的大小排，每次从最小的开始替换，如果最小的值比当前的$C$大就可以不用换了。这样保证了每个数最多进出队一次。

时间复杂度$O\left( n\log \left( n\right) \right)$(对吗?)

#include 
#define ll long long
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

const int M = 1e5 + 10;

struct P {
    int b;ll c;
    bool operator < (const P &rhs) const {
        return c > rhs.c;      
    }
} p[M];

int main() {
    int n = read(), m = read();
    priority_queue, greater > que;
    for (int i = 0; i < n; i++) {
        int x = read();
        que.push((ll)x);
    }    
    for (int i = 0; i < m; i++) p[i].b = read(), p[i].c = (ll)read();
    sort(p, p + m);
    for (int i = 0; i < m; i++) {
        int b = p[i].b;
        if (p[i].c <= que.top()) break;
        while (b--) {
            if (que.top() < p[i].c) {
                que.pop();
                que.push(p[i].c);
            } else {
                break;
            }
        }
    }
    ll sum = 0;
    while (!que.empty()) {
        int x = que.top(); que.pop();
        sum += x;
    }
    cout << sum << '\n';
    return 0;
}

View Code

 

E - Cell Distance

题意：求一个网格图里面任取$K$点的曼哈顿距离之和

思路：$N\times M$的网格图里面任意两点的曼哈顿距离的平均值是$\dfrac {N+M}{3}$

答案就是$C^{k}_{N\times M}C^{2}_{k}\dfrac {N+M}{3}$

#include 
#define ll long long
using namespace std;

const ll MOD = 1e9 + 7;

inline ll read() {
    ll x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

ll qp(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

const int N = 2e5 + 10;
ll fac[N];

ll C(ll a, ll b) {
    ll ans = fac[a] * qp(fac[b], MOD - 2) % MOD;
    ans = ans * qp((fac[a-b] + MOD) % MOD, MOD - 2) % MOD;
    return ans;
}

int main() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i % MOD;
    ll n = read(), m = read(), k = read();
    ll ans = C(n * m, k) * C(k, 2) % MOD * qp(3, MOD - 2) % MOD;
    ans = ans * (n + m) % MOD;
    ans %= MOD;
    cout << ans << '\n';
    return 0;
}

View Code

 

F - Absolute Minima

题意：有函数$f\left( x\right)$初始为0，两个操作，1是给$f\left( x\right)$加上$\left| x-a\right| +b$，2是询问函数的最小值及$x$

思路：可以证(举例)明(发现)，函数的最小值一定$a$序列的中位数取到。

两个优先队列，一个降序一个升序，每次加入$a$都加入两个队列里，在比较他们的顶，降序的顶必须小于等于升序的顶，这样就能实现两个堆分别存储了序列的左半部分和右半部分。

同时，降序的顶就是取到的$x$

#include 
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int main() {
    int q = read();
    priority_queue<int> l;
    priority_queue<int, vector<int>, greater<int> > r;
    long long ans = 0;
    while (q--) {
        int t = read();
        if (t == 1) {
            int a = read(), b = read();
            ans += b;
            l.push(a);
            r.push(a);
            if (l.top() > r.top()) {
                int x = l.top(); l.pop();
                int y = r.top(); r.pop();
                ans += abs(x - y);
                l.push(y); r.push(x);
            }
        } else {
            printf("%d %lld\n", l.top(), ans);
        }
    }
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/Mrzdtz220/p/10937591.html