标签：prime AtCoder ch frac Beginner 300 ++ int &&

A - N-choice question

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 1e5+5;
bool vis[N];
vector<int> e[N] ;
int d[N];

int main() {
	int n = read() , a = read() , b = read();
    for( int x , i = 1 ; i <= n ; i ++ ){
        x = read();
        if( x == a + b ) cout << i , exit(0);
    }
    
    return 0;
}

B - Same Map in the RPG World

看起来似乎很复杂，其实可以直接枚举横着移动多少次，竖着移动多少次就行了。

#include<bits/stdc++.h>

using namespace std;


int main() {
	ios::sync_with_stdio(false) , cin.tie(nullptr) , cout.tie(nullptr);
    int n , m;
    cin >> n >> m;
    vector<string> a(n) , b(n);
    for( auto & i : a ) cin >> i;
    for( auto & i : b ) cin >> i;
    for( int r = 0 ; r < n ; r ++ )
        for( int c = 0 ; c < m ; c ++ ){
            int f = 1;
            for( int i = 0 ; f && i < n ; i ++ )
                for( int j = 0 ; f && j < m ; j ++ ){
                    if( a[i][j] != b[(i+r)%n][ (j+c) % m ] ) f = 0; 
                }
            if( f ) cout << "Yes\n" , exit(0); 
        }
    cout << "No\n";
    return 0;
}

C - Cross

暴力枚举点的坐标，暴力枚举大小，然后暴力判断即可。

#include<bits/stdc++.h>

using namespace std;

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

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    vector<int> cnt(min(n, m) + 1);
    for (auto &i: g) cin >> i;

    for (int i = 0, t; i < n; i++)
        for (int j = 0; j < m; j++) {
            if (g[i][j] == '.') continue;
            t = 0;
            for (int k = 1, f; k; k++) {
                f = 1;
                for (int l = 0; l < 4 && f; l++)
                    if (g[i + dx[l] * k][j + dy[l] * k] == '.') f = 0;
                if (f) t = k;
                else break;
            }
            cnt[t]++;
        }
    for (int i = 1; i < cnt.size(); i++) cout << cnt[i] << " ";
    return 0;
}

D - AABCC

首先求一下素数，然后枚举\(a,b\)二分\(c\)就好了。

#include<bits/stdc++.h>

using namespace std;

#define int __int128
typedef long long ll;


int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    const int N = 300000;
    vector<int> not_prime(N + 1, 0);
    vector<int> prime;
    not_prime[0] = not_prime[1] = true;
    for (int i = 2; i <= N; i++) {
        if (not_prime[i]) continue;
        prime.push_back(i);
        for (int j = i * 2; j <= N; j += i)
            not_prime[j] = 1;
    }
    int n = read(), res = 0;
    for (int i = 0, a, b; i < prime.size(); i++) {
        a = prime[i] * prime[i];
        if (a > n) break;
        for (int j = i + 1; j < prime.size(); j++) {
            b = prime[j] * a;
            if (b > n) break;
            int l = j + 1, r = prime.size() - 1, mid, ans = -1;
            while (l <= r) {
                mid = (l + r) >> 1;
                if (b * prime[mid] * prime[mid] <= n) ans = mid, l = mid + 1;
                else r = mid - 1;
            }
            if (ans == -1) break;
            res += ans - j;
        }

    }
    cout << (ll) res << "\n";
    return 0;

}

E - Dice Product 3

有题可知

\(f(i)=\frac 1 6(f(i)+f(\frac i 2 ) + f(\frac i 3 ) + f(\frac i 4) + f(\frac i 5) + f(\frac i 6) )\)

\(\frac 5 6f(i)=\frac 1 6(f(\frac i 2 ) + f(\frac i 3 ) + f(\frac i 4) + f(\frac i 5) + f(\frac i 6) )\)

\(f(i)=\frac 1 5(f(\frac i 2 ) + f(\frac i 3 ) + f(\frac i 4) + f(\frac i 5) + f(\frac i 6) )\)

其中\(f(1)=1\)，如果不能整除的话就是\(0\)

然后直接记忆化搜索一下？

#include<bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int mod = 998244353, inv = 598946612;

map<int, int> f;

int dp(int n) {
    if (f.count(n)) return f[n];
    int ans = 0;
    for (int i = 2; i <= 6; i++)
        if (n % i == 0) ans = (ans + dp(n / i)) % mod;
    return f[n] = ans * inv % mod;
}

int32_t main() {
    int n;
    cin >> n;
    f[1] = 1;
    cout << dp(n);
    return 0;
}

F - More Holidays

记录\(S\)中每一个x的位置，很容易可以想到相邻的x一定是最优的，所以直接枚举一下起点就好。但是枚举的是时候实际上只用枚举第一个\(S\)中的位置就可以。然后特判第一个和最后一个的特殊情况。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main(){
	ios::sync_with_stdio(false) , cin.tie(nullptr) , cout.tie(nullptr);
	int n , m , k;
	string s;
	cin >> n >> m >> k >> s;
	vector<int> p;
	for( int i = 0 ; i < n ; i ++ )
		if( s[i] == 'x' ) p.push_back(i);
	int cnt = p.size() , t = cnt * m;
	if( cnt * m == k ) cout << n * m , exit(0);
	int res = 0;
	res = max( res , n * ( k / cnt ) + p[k%cnt] );
	res = max( res , n * ( k / cnt ) + n - 1 - p[cnt - 1 - k % cnt] );
	for( int i = 0 ; i < cnt ; i ++ ){
		if( k + i + 1 < t )
			res = max( res , p[(k+i+1)%cnt] + n * ((k+i+1) / cnt) - p[i] - 1 );
	}
	cout << res;
	return 0; 
}

标签：prime,AtCoder,ch,frac,Beginner,300,++,int,&&
From： https://www.cnblogs.com/PHarr/p/17369742.html

AtCoder Beginner Contest 300