Caddi Programming Contest 2021(AtCoder Beginner Contest 193) 题解

Caddi Programming Contest 2021(AtCoder Beginner Contest 193)

A - Discount

打折浮点数除即可

B - Play Snuke

枚举判断符合要求的求最小值即可

C - Unexpressed

$O(\sqrt{n})$枚举$a$，暴力翻倍（最小的$2$最多乘$32$次就会超过$n$的上限）

$map$存该数有没有被$a^b$覆盖到，数量不会太多，不用担心$MLE$

D - Poker

暴力枚举，计算情况。注意两个串选择同一数字时的小细节即可

#include <cstdio>
#include <cmath>
#define int long long
#define maxn 20
int n;
char s[maxn], t[maxn];
int tot[maxn], cnt_s[maxn], cnt_t[maxn];

int calc( int x ) {
	return x * ( x - 1 );
}

signed main() {
	scanf( "%lld %s %s", &n, s + 1, t + 1 );
	for( int i = 1;i <= 4;i ++ )
		tot[s[i] - '0'] ++, tot[t[i] - '0'] ++;
	int cnt = 0;
	for( int i = 1;i <= 9;i ++ ) {
		if( tot[i] == n ) continue;
		for( int j = 1;j <= 9;j ++ ) {
			if( ( tot[j] == n ) || ( i == j && tot[i] + 1 == n ) ) continue;
			int sum_s = 0, sum_t = 0;
			s[5] = i + '0', t[5] = j + '0';
			for( int k = 1;k <= 5;k ++ )
				cnt_s[s[k] - '0'] ++, cnt_t[t[k] - '0'] ++;
			for( int k = 1;k <= 9;k ++ ) {
				sum_s += k * pow( 10, cnt_s[k] );
				sum_t += k * pow( 10, cnt_t[k] );
			}
			if( sum_s > sum_t ) {
				if( i == j )
					cnt += ( n - tot[i] ) * ( n - tot[i] - 1 );
				else
					cnt += ( n - tot[i] ) * ( n - tot[j] );
			}
			for( int k = 1;k <= 9;k ++ )
				cnt_s[k] = cnt_t[k] = 0;
		}
	}
	printf( "%.10f\n",  cnt * 1.0 / ( calc( 9 * n - 8 ) ) );
	return 0;
}
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E - Oversleeping

发现每段差都在500以内，暴力枚举相遇点

$k_1\ast (2X+2Y)+X+i=k_2\ast (P+Q)+P+j\iff k_1\ast (2X+2Y)-k_2\ast (P+Q)=P-X+j-i$

即扩欧，$k_1^{\prime }(2X+2Y)+k_2^{\prime }(P+Q)=gcd(2X+2Y,P+Q)$

有解必须满足$gcd(2X+2Y,P+Q)|P-X+j-i$

扩欧求出的解不一定是非负最优解，扩倍(乘对方的倍数)成非负再取模变最优

#include <cstdio>
#include <iostream>
using namespace std;
#define int __int128
int T, X, Y, P, Q;

void read( int &x ) {
	x = 0; char s = getchar();
	while( s < '0' || s > '9' ) s = getchar();
	while( '0' <= s && s <= '9' ) {
		x = ( x << 1 ) + ( x << 3 ) + ( s ^ 48 );
		s = getchar();
	}
}

void exgcd( int a, int b, int &d, int &x, int &y ) {
	if( ! b ) d = a, x = 1, y = 0;
	else {
		exgcd( b, a % b, d, y, x );
		y -= x * ( a / b );
	}
}

void print( int x ) {
	if( x > 9 ) print( x / 10 );
	putchar( ( x % 10 ) + '0' );
}

signed main() {
	read( T );
	while( T -- ) {
		read( X ), read( Y ), read( P ), read( Q );
		int d, x, y, ans = 1e32;
		exgcd( X * 2 + Y * 2, P + Q, d, x, y );
		for( int i = 0;i < Y;i ++ )
			for( int j = 0;j < Q;j ++ ) {
				int c = P + j - X - i;
				if( c % d ) continue;
				int k1 = c / d * x;
				int k2 = c / d * ( - y );
				int mod1 = ( P + Q ) / d, mod2 = ( X * 2 + Y * 2 ) / d;
				if( k1 < 0 ) {
					int t = ( - k1 + mod1 - 1 ) / mod1;
					k1 += mod1 * t, k2 += mod2 * t;
				}
				if( k2 < 0 ) {
					int t = ( - k2 + mod2 - 1 ) / mod2;
					k1 += mod1 * t, k2 += mod2 * t;
				}
				k1 %= mod1;
				ans = min( ans, k1 * ( X * 2 + Y * 2 ) + X + i );
			}
		if( ans == 1e32 ) printf( "infinity\n" );
		else print( ans ), putchar( '\n' );
	}
	return 0;
}
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F - Zebraness

一般的这种相邻之间计算贡献都可以转换成最小割问题，考虑整体边数减去最小失去

如果相邻两个点是一样的颜色，则失去$1$

但是这样根本无法在网络流上进行切割

所以考虑奇偶分开，天然分裂相邻块

翻转奇数块颜色，进行连边；固定颜色的最大可以流出$4$，至于不确定的就根据相邻边选择

按颜色分给超级源点和超级汇点，大约是在$O(n^3)$

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 105
#define inf 0x7f7f7f7f
struct node {
	int v, nxt, flow;
}edge[maxn * maxn * 30];
int n, cnt, s, t;
char ch[maxn][maxn];
int head[maxn * maxn], cur[maxn * maxn], dis[maxn * maxn];
queue < int > q;

void addEdge( int u, int v, int c ) {
	edge[cnt].v = v;
	edge[cnt].nxt = head[u];
	edge[cnt].flow = c;
	head[u] = cnt ++;
	edge[cnt].v = u;
	edge[cnt].nxt = head[v];
	edge[cnt].flow = 0;
	head[v] = cnt ++;
}

int id( int i, int j ) {
	return ( i - 1 ) * n + j;
}

bool bfs() {
	memset( dis, 0, sizeof( dis ) );
	memcpy( cur, head, sizeof( head ) );
	dis[s] = 1; q.push( s );
	while( ! q.empty() ) {
		int u = q.front(); q.pop();
		for( int i = head[u];~ i;i = edge[i].nxt ) {
			int v = edge[i].v;
			if( ! dis[v] && edge[i].flow ) {
				dis[v] = dis[u] + 1;
				q.push( v );
			}
		}
	}
	return dis[t];
}

int dfs( int u, int cap ) {
	if( u == t || ! cap ) return cap;
	int flow = 0;
	for( int i = cur[u];~ i;i = edge[i].nxt ) {
		int v = edge[i].v; cur[u] = i;
		if( dis[v] == dis[u] + 1 ) {
			int w = dfs( v, min( cap, edge[i].flow ) );
			if( ! w ) continue;
			cap -= w;
			flow += w;
			edge[i].flow -= w;
			edge[i ^ 1].flow += w;
			if( ! cap ) break;
		}
	}
	return flow;
}

int dinic() {
	int ans = 0;
	while( bfs() ) ans += dfs( s, inf );
	return ans;
}

int main() {
	memset( head, -1, sizeof( head ) );
	scanf( "%d", &n );
	for( int i = 1;i <= n;i ++ )
		scanf( "%s", ch[i] + 1 );
	for( int i = 1;i <= n;i ++ )
		for( int j = 1;j <= n;j ++ )
			if( ! ( ( i + j ) & 1 ) || ch[i][j] == '?' ) continue;
			else ch[i][j] = 'B' + 'W' - ch[i][j];
	s = 0, t = n * n + 1;
	for( int i = 1;i <= n;i ++ )
		for( int j = 1;j <= n;j ++ ) {
			if( ch[i][j] == 'B' ) addEdge( s, id( i, j ), 4 );
			if( ch[i][j] == 'W' ) addEdge( id( i, j ), t, 4 );
		}
	for( int i = 1;i <= n;i ++ )
		for( int j = 1;j < n;j ++ ) {
			addEdge( id( i, j ), id( i, j + 1 ), 1 );
			addEdge( id( i, j + 1 ), id( i, j ), 1 );
		}
	for( int i = 1;i < n;i ++ )
		for( int j = 1;j <= n;j ++ ) {
			addEdge( id( i, j ), id( i + 1, j ), 1 );
			addEdge( id( i + 1, j ), id( i, j ), 1 );
		}
	printf( "%d\n", 2 * n * ( n - 1 ) - dinic() );
	return 0;
}
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