atcoder题目链接

A - Your First Judge

签到题

#include <cstdio>
#include <iostream>
using namespace std;
string s;
int main() {cin >> s;if( s == "Hello,World!" ) printf( "AC\n" );else printf("WA\n" );return 0;
}

B - log2(N)

签到题

#include <cstdio>
int main() {long long n;scanf( "%lld", &n );int ans;for( int i = 0;;i ++ ) {if( ( 1ll << i ) <= n ) ans = i;else break;}printf( "%d", ans );return 0;
}

C - One More aab aba baa

离散化后压成整数搜索，简单题

#include <stack>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
stack < char > sta;
char s[10], rnk[10];
int t[40325], used[30];
bool vis[10];
int n, k, cnt, tot;void dfs( int x, int val ) {if( x > n ) {t[++ cnt] = val;return;}for( int i = 1;i <= n;i ++ )if( vis[i] ) continue;else {vis[i] = 1;dfs( x + 1, val * 10 + used[s[i] - 'a' + 1] );vis[i] = 0;}
}int main() {scanf( "%s %d", s + 1, &k );n = strlen( s + 1 );for( char i = 'a';i <= 'z';i ++ )for( int j = 1;j <= n;j ++ )if( s[j] == i ) {used[i - 'a' + 1] = ++ tot, rnk[tot] = i;break;}dfs( 1, 0 );sort( t + 1, t + cnt + 1 );cnt = unique( t + 1, t + cnt + 1 ) - t - 1;while( t[k] ) sta.push( rnk[t[k] % 10] ), t[k] /= 10;while( ! sta.empty() ) printf( "%c", sta.top() ), sta.pop();return 0;
}

D - Coprime 2

质因数分解后，把质因数的倍数都筛掉，简单题

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define maxn 100005
vector < int > ans;
int cnt;
bool vis[maxn], is[maxn];
int prime[maxn];void init( int n ) {for( int i = 2;i <= n;i ++ ) {if( ! vis[i] ) is[i] = 1, prime[++ cnt] = i;for( int j = 1;j <= cnt && i * prime[j] <= n;j ++ ) {vis[i * prime[j]] = 1;if( i % prime[j] == 0 ) break;}}
}int n, m;
int a[maxn], p[maxn * 100];int main() {init( 1e5 );scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] );cnt = 0;for( int i = 1;i <= n;i ++ ) {for( int j = 1;j * j <= a[i];j ++ )if( a[i] % j == 0 ) {if( is[j] ) p[++ cnt] = j;if( is[a[i] / j] ) p[++ cnt] = a[i] / j;}}for( int i = 1;i <= m;i ++ ) vis[i] = 0;sort( p + 1, p + cnt + 1 );cnt = unique( p + 1, p + cnt + 1 ) - p - 1;for( int i = 1;i <= cnt;i ++ )for( int j = p[i];j <= m;j += p[i] )vis[j] = 1;for( int i = 1;i <= m;i ++ )if( ! vis[i] ) ans.push_back( i );printf( "%d\n", ans.size() );for( int i = 0;i < ans.size();i ++ )printf( "%d\n", ans[i] );return 0;
}

E - Chain Contestant

只有$10$种，符合状压

$d{p}_{i,s,j}:$ 到$i$为止出现字符的状态为$s$，上一次出现的字符为$j$（多加一种字符表示空）

• 不选$i$

  $d{p}_{i,s,j}+=dp[i-1][s][j]$

• 选$i$，必须满足$i$位置字符没出现过或上一次出现字符就是$i$位置字符

  $dp[i][1<<t[i]|s][t[i]]+=dp[i][1<<t[i]|s][t[i]]+dp[i-1][s][j]$

简单题

#include <cstdio>
#define int long long
#define mod 998244353
#define maxn 1002
#define S 1 << 10
int n;
char s[maxn];
int t[maxn];
int dp[maxn][S][11];signed main() {scanf( "%lld %s", &n, s + 1 );for( int i = 1;i <= n;i ++ ) t[i] = s[i] - 'A';dp[0][0][10] = 1;for( int i = 1;i <= n;i ++ )for( int sta = 0;sta < S;sta ++ )for( int j = 0;j <= 10;j ++ )if( ! dp[i - 1][sta][j] ) continue;else {dp[i][sta][j] = ( dp[i][sta][j] + dp[i - 1][sta][j] ) % mod;if( j == 10 ) dp[i][1 << t[i] | sta][t[i]] = ( dp[i][1 << t[i] | sta][t[i]] + dp[i - 1][sta][j] ) % mod;else if( ( 1 << t[i] & sta ) and j != t[i] ) continue;elsedp[i][1 << t[i] | sta][t[i]] = ( dp[i][1 << t[i] | sta][t[i]] + dp[i - 1][sta][j] ) % mod;}int ans = 0;for( int i = 0;i < S;i ++ )for( int j = 0;j <= 10;j ++ )if( i == 0 && j == 10 ) continue;else ans = ( ans + dp[n][i][j] ) % mod;printf( "%lld\n", ans );return 0;
}

F - Dist Max 2

二分答案

将点按$x$坐标排序

类似滑动窗口模拟

记录在满足$x$坐标情况下的最大最小$y$，判断是否有满足与现在枚举$i$的$y$距离不小于二分答案

简单题

#include <queue>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 200005
struct node { int x, y; }p[maxn];
queue < node > q;
int n;int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) scanf( "%d %d", &p[i].x, &p[i].y );sort( p + 1, p + n + 1, []( node s, node t ) { return s.x < t.x; } );int l = 0, r = 1e9, ans;next :while( l <= r ) {int mid = l + r >> 1;while( ! q.empty() ) q.pop();int Min = 1e9, Max = 0;for( int i = 1;i <= n;i ++ ) {while( ! q.empty() ) {if( q.front().x > p[i].x - mid ) break;Min = min( Min, q.front().y );Max = max( Max, q.front().y );q.pop();}if( Min <= p[i].y - mid or Max >= p[i].y + mid ) {ans = mid, l = mid + 1;goto next;}q.push( p[i] );}r = mid - 1;}printf( "%d\n", ans );return 0;
}

G - Colorful Candies 2

中档题

令$C$表示不同的糖果颜色数，将$N$颗糖果重新离散化颜色1,2,...,C

显然答案与具体糖果的颜色无关，只与每种颜色的个数有关，令$n_i$表示颜色为$i$的糖果个数

枚举每一轮要选择的糖果个数$K$

对于每一种颜色，定义$X_i$，若选择糖果种有$i$颜色，$X_i=1$，否则$X_i=0$

则$K$糖果中不同颜色种数可以表示为${\sum }_{i=1}^C{X}_i$

所求期望为$E\left[{\sum }_{i=1}^C{X}_i\right]$，根据期望的线性可加性知道$={\sum }_{i=1}^{C}E\left[X_i\right]$

一个颜色出现的期望很简单，出现的方案数$/$总方案数，出现的方案数$=$总方案数$-$一次都没出现
$$E\left[X_i\right]=\frac{\left(\genfrac{}{}{0ex}{}{N}{K}\right)-\left(\genfrac{}{}{0ex}{}{N-n_i}{K}\right)}{\left(\genfrac{}{}{0ex}{}{N}{K}\right)}$$
枚举$i$又是$O(n)$的时间，再加上外层的$K$枚举，时间$O(n^2)$难以接受

发现其实如果有$x$种颜色的个数都是$n_i$，那么这$x$种颜色的期望可以$O(1)$计算
$$x\ast \frac{\left(\genfrac{}{}{0ex}{}{N}{K}\right)-\left(\genfrac{}{}{0ex}{}{N-n_i}{K}\right)}{\left(\genfrac{}{}{0ex}{}{N}{K}\right)}$$
那么不同个数就是$n_i=1,2,...,$，又要满足$\sum n_i=N$，所以可以知道上界就是根号级别$n_i=1,2,...,\sqrt{N}$

#include <map>
#include <cstdio>
using namespace std; 
#define int long long 
#define mod 998244353
#define maxn 50005
map < int, int > color, cnt;
int fac[maxn], inv[maxn];
int n;int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= n;i ++ ) fac[i] = fac[i - 1] * i % mod;inv[n] = qkpow( fac[n], mod - 2 );for( int i = n - 1;i;i -- ) inv[i] = inv[i + 1] * ( i + 1 ) % mod;
}int C( int n, int m ) {if( n < m ) return 0;else return fac[n] * inv[m] % mod * inv[n - m] % mod;
}signed main() {scanf( "%lld", &n );init();for( int i = 1, x;i <= n;i ++ ) {scanf( "%lld", &x );color[x] ++;//颜色x的出现次数 }for( auto it : color )	cnt[it.second] ++;//颜色出现次数为x的颜色个数for( int i = 1;i <= n;i ++ ) {int ans = 0;for( auto it : cnt ) ans = ( ans + ( C( n, i ) - C( n - it.first, i ) + mod ) % mod * it.second ) % mod;printf( "%lld\n", ans * qkpow( C( n, i ), mod - 2 ) % mod );}return 0;
}

后言：$H$就是$FWT$，atcoder-abc很喜欢在最后考些鹅心算法模板，不想补了