AtCoder Beginner Contest 166 题解

比赛的时候只切了 (5) 题，自闭了。

赛后抱着试一试的心态 F 交了一发 dfs，结果就过了？？？

看来比赛的时候还是要敢写敢交啊……

不过还好上青了……

A - A?C

题面

简单字符串。

#include <bits/stdc++.h>

using namespace std;

const int N = 200003, M = N << 1;

int n, k;
int a[N], b[N];
char s[N];

int main()
{
    scanf("%s", s + 1);
    if (s[2] == 'R') s[2] = 'B';
    else s[2] = 'R';
    cout << s + 1 << endl;
    return 0;
}

B - Trick or Treat

题面

简单模拟，开个桶即可。

#include <bits/stdc++.h>

using namespace std;

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

const int N = 103, M = N << 1;

int n, k;
int a[N], b[N];

int main()
{
    n = gi(), k = gi();
    for (int i = 1; i <= k; i+=1)
    {
        int d = gi();
        for (int j = 1; j <= d; j+=1)
            a[j] = gi(), ++b[a[j]];
    }
    int cnt = 0;
    for (int i = 1; i <= n; i+=1) cnt += (b[i] == 0);
    cout << cnt << endl;
    return 0;
}

C - Peaks

题面

建完图后枚举每个点判断是否满足条件。

#include <bits/stdc++.h>

using namespace std;

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

const int N = 100003, M = N << 1;

int n, k;
int a[N];
int tot, head[N], ver[M], nxt[M];

inline void add(int u, int v)
{
    ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

int main()
{
    n = gi(), k = gi();
    for (int i = 1; i <= n; i+=1) a[i] = gi();
    for (int i = 1; i <= k; i+=1)
    {
        int u = gi(), v = gi();
        add(u, v), add(v, u);
    }
    int cnt = 0;
    for (int i = 1; i <= n; i+=1)
    {
        bool fl = true;
        for (int j = head[i]; j; j = nxt[j])
        {
            int v = ver[j];
            if (a[v] >= a[i]) {fl = false; break;}
        }
        if (fl) ++cnt;
    }
    cout << cnt << endl;
    return 0;
}

D - I hate Factorization

题面

一开始以为是一道因式分解的数学题，结果发现只需要确定一个范围后枚举就行了。

我把 (A) 和 (B) 的范围限定在 ([-7777,7777])，这样也可以过。

预处理负数的时候可以使用数组偏移的技巧。

注意开 ( ext{long long})。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

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

const int N = 100003, M = N << 1;

int n, k;
LL fac5[N];

int main()
{
    n = gi();
    LL now = 1;
    for (int i = 1; i <= 20000; i+=1) 
    {
        now = 1;
        for (int j = 1; j <= 5; j+=1) now *= (i - 7778);
        fac5[i] = now;
    }
    for (int i = -7777; i <= 7777; i+=1)
        for (int j = -7777; j <= 7777; j+=1)
            if (fac5[i + 7778] - fac5[j + 7778] == n)
            {
                printf("%d %d
", i, j);
                return 0;
            }
    return 0;
}

E - This Message Will Self-Destruct in 5s

题面

我们假设选的两个数分别是 (i) 和 (j) ((i>j))。

根据题意可得：(i-j=a_i+a_j)。

移项：(i-a_i=j+a_j)。

那么我们对于每一个 (i)，开个桶记录一下区间 ([1, i-1]) 中等于 (i-a_i) 的 (j+a_j) 的个数 ((jin [1, i-1]))。

注意到 (i-a_i) 最大为 (199999)，因此我们只需要记录 (leq 199999) 的 (j+a_j) 的个数。

倒序循环计算答案即可。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

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

const int N = 200003, M = N << 1;

int n, k;
int a[N], b[N];
int s[N] /*i + a[i]*/, t[N] /*j - a[j]*/;
int ton[M];
LL ans;

int main()
{
    n = gi();
    for (int i = 1; i <= n; i+=1) 
        a[i] = gi(), s[i] = i + a[i], t[i] = i - a[i];
    for (int i = 1; i <= n; i+=1)
        if (s[i] < 200000) ++ton[s[i]];
    for (int i = n; i >= 1; i-=1)
    {
        if (t[i] <= 0) continue;
        if (t[i] >= 200000) continue;
        ans += ton[t[i]];
        if (s[i] < 200000) --ton[s[i]];
    }
    printf("%lld
", ans);
    return 0;
}

F - Three Variables Game

题面

本来以为很复杂，要很多种分类讨论，于是比赛时一直没写出来。

赛后看见有人用 dfs 过了，于是我试着也写一写 dfs。

然后就过了？？？

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

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

const int N = 200003, M = N << 1;

int n, k;
int a, b, c;
string s[N];
string ans[N];

void dfs(int now, int a, int b, int c)
{
    if (now > n) 
    {
        puts("Yes");
        for (int i = 1; i <= n; i+=1) cout << ans[i] << endl;
        exit(0);
    }
    if (s[now] == "AB")
    {
        if (b > 0) ans[now] = "A", dfs(now + 1, a + 1, b - 1, c);
        if (a > 0) ans[now] = "B", dfs(now + 1, a - 1, b + 1, c);
    }
    else if (s[now] == "AC")
    {
        if (c > 0) ans[now] = "A", dfs(now + 1, a + 1, b, c - 1);
        if (a > 0) ans[now] = "C", dfs(now + 1, a - 1, b, c + 1);
    }
    else 
    {
        if (c > 0) ans[now] = "B", dfs(now + 1, a, b + 1, c - 1);
        if (b > 0) ans[now] = "C", dfs(now + 1, a, b - 1, c + 1);
    }
}

int main()
{
    n = gi(), a = gi(), b = gi(), c = gi();
    for (int i = 1; i <= n; i+=1) cin >> s[i];
    dfs(1, a, b, c);
    puts("No");
    return 0;
}