比赛链接

A - Signed Difficulty

签到题

#include <cstdio>
int x, y;
char c;
int main() {scanf( "%d%c%d", &x, &c, &y );if( y <= 2 ) printf( "%d-", x );else if( y >= 7 ) printf( "%d+", x );else printf("%d", x );return 0;
}

B - Same Name

签到题

#include <cstdio>
#include <iostream>
#include <map>
using namespace std;
map < pair < string, string >, int > mp;
int n;int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) {string a, b;cin >> a >> b;if( mp[make_pair( a, b )] ) return ! printf( "Yes\n" );mp[make_pair( a, b )] = 1;}printf( "No\n" );return 0;
}

C - Many Balls

简单题

发现是快速幂的逆过程

#include <cstdio>
#include <vector>
using namespace std;
#define int long long
vector < int > ans;
int n;signed main() {scanf( "%lld", &n );while( n ) {if( n & 1 ) ans.push_back( 0 );ans.push_back( 1 );n >>= 1;}for( int i = ans.size() - 1;~ i;i -- )printf( "%c", ans[i] + 'A' );return 0;
}

D - Pair of Balls

超难题！！！靠

不搞环，硬汉就搞大模拟

记录颜色两次出现位置，并带上标记，成为堆顶为$0$，在下面为$1$

当颜色两个位置都入队（成为其所在堆的堆顶）

弹出，并把堆下面一个元素标记为新的堆顶

就这么模拟，凸(艹皿艹 )

之前打的什么鬼傻逼set/map半天模拟不动，又TLE又RE又WA

o(╥﹏╥)o _(¦3」∠)_ ┏┛墓┗┓…(((m-__-)m

#include <cstdio>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
#define maxn 200005
struct node {int color, i, rnk, pos;node(){}node( int C, int I, int R, int P ) {color = C, i = I, rnk = R, pos = P;}
}c[maxn][2];
int n, m, cnt;
queue < node > q;
vector < int > g[maxn];
bool vis[maxn][2];int main() {scanf( "%d %d", &n, &m );for( int i = 1, k, x;i <= m;i ++ ) {scanf( "%d", &k );for( int j = 1;j <= k;j ++ ) {scanf( "%d", &x );g[i].push_back( x );if( ! vis[x][0] ) c[x][0] = node( x, i, j, 1 ), vis[x][0] = 1;else c[x][1] = node( x, i, j, 1 ), vis[x][1] = 1;}}for( int i = 1;i <= n;i ++ ) {auto ls = c[i][0];auto rs = c[i][1];vis[i][0] = vis[i][1] = 0;if( ! ls.pos and ! rs.pos ) {q.push( c[i][0] );q.push( c[i][1] );}}while( ! q.empty() ) {auto ls = q.front(); q.pop();auto rs = q.front(); q.pop();cnt ++;int nxt;if( ls.rnk ^ g[ls.i].size() ) {nxt = g[ls.i][ls.rnk];if( c[nxt][0].i == ls.i ) c[nxt][0].pos = 0;else c[nxt][1].pos = 0;if( ! c[nxt][0].pos and ! c[nxt][1].pos ) {q.push( c[nxt][0] );q.push( c[nxt][1] );}}if( rs.rnk ^ g[rs.i].size() ) {nxt = g[rs.i][rs.rnk];if( c[nxt][0].i == rs.i ) c[nxt][0].pos = 0;else c[nxt][1].pos = 0;if( ! c[nxt][0].pos and ! c[nxt][1].pos ) {q.push( c[nxt][0] );q.push( c[nxt][1] );}}}if( cnt ^ n ) printf( "No\n" );else printf( "Yes\n" );return 0;
}

E - Amusement Park

简单题

数学直接算

从大到小取，每次$a[i]$取到$a[i+1]$相同值就停，等差数列算这中间的值和

随着$i$的枚举，前面全是相同的$a[i]$，所以这等差数列算了后还要乘以一个$i$的系数

#include <cstdio>
#include <algorithm>
using namespace std;
#define int long long
#define maxn 100005
int n, k;
int a[maxn];signed main() {scanf( "%lld %lld", &n, &k );for( int i = 1;i <= n;i ++ )scanf( "%lld", &a[i] );sort( a + 1, a + n + 1, []( int x, int y ) { return x > y; } );int ans = 0;for( int i = 1;i <= n;i ++ ) {int x = a[i] - a[i + 1];int sum = ( a[i] + a[i + 1] + 1 ) * x / 2;if( k <= x * i ) {x = k / i;sum = ( a[i] + a[i] - x + 1 ) * x / 2;ans += sum * i;a[i] -= k / i;k %= i;ans += a[i] * k;break;}else {ans += sum * i;k -= x * i;}}printf( "%lld\n", ans );return 0;
}

F - Max Sum Counting

简单题

凸(艹皿艹 )，傻逼背包倒DP调nm半个小时

按$A$排序，枚举$A_i$作为最大值

接下来就是选一些数满足$B_j$的和不超过$A_i$

设$d{p}_j:$ 和为$j$的方案数

每次统计答案后，再转移$i$的$B_i$的贡献

发现其实就是个简单背包

#include <cstdio>
#include <algorithm>
using namespace std;
#define int long long
#define mod 998244353
#define maxn 5005
struct node { int a, b; }g[maxn];
int n;
int t[maxn], dp[maxn];bool cmp( node x, node y ) {return x.a < y.a;
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &g[i].a );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &g[i].b );sort( g + 1, g + n + 1, cmp );int ans = 0;dp[0] = 1;for( int i = 1;i <= n;i ++ ) {for( int j = 0;j <= g[i].a - g[i].b;j ++ )ans = ( ans + dp[j] ) % mod;for( int j = maxn - 1;j >= g[i].b;j -- )dp[j] = ( dp[j] + dp[j - g[i].b] ) % mod;}printf( "%lld\n", ans );return 0;
}

G - 01Sequence

看完题目是非常板的差分约束题面感

设$s_i$表示$1-i$中1的个数

$[l,r]$区间内1个数不少于$x$个，翻译约束为$s_r-s_{l-1}\ge x$

为了契合差分约束的$\le$

重新定义$s_i$表示$1-i$中0的个数

翻译约束为$s_r-s_{l-1}\le r-l+1-x$

转化到图论上，连边$l-1\to r$有向边的边权$r-l+1-x$

跑最短路即可

本题卡SPFA

差分约束

• 为什么是$\le$约束（不等式左边为“点”，不等式右边为固定限制）

  • 首先最长路目前没有快速地算法
  • 其次符合最短路$di{s}_v\le di{s}_u+w_{u,v}$的形式

• 感觉最后求$s_n$的最大值，应该是最长路，为什么图论上是跑最短路

  • 首先约束和$s_n$之间是逆向关系，并不是正比于

  • 可以理解为最短路就是求出最严格的约束

  • $s_n$的所有值都保证要满足所有的约束，自然要满足最严格的约束

  • e.g. $(i,j)=x_1,(j,k)=x_2,(i,k)=x_3,x_1+x_2<x_3$

    既要满足$di{s}_j\le di{s}_i+x_1;di{s}_k\le di{s}_j+x_2$

    又要满足$di{s}_k\le di{s}_i+x_3$

    并且$di{s}_k\le di{s}_j+x_2\le di{s}_i+x_1+x_2\le di{s}_i+x_3$

    根据不等式的原则，同小取小的最严格约束

    这恰好匹配最短路的过程

所以差分约束的本质是转化为最短路问题

#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 200005
#define Pair pair < int, int >
priority_queue < Pair, vector < Pair >, greater < Pair > > q;
int n, m;
int dis[maxn];
bool vis[maxn];
vector < Pair > G[maxn];void addedge( int u, int v, int w ) {G[u].push_back( make_pair( v, w ) );
}int main() {scanf( "%d %d", &n, &m );for( int i = 1, l, r, x;i <= m;i ++ ) {scanf( "%d %d %d", &l, &r, &x );addedge( l - 1, r, r - l + 1 - x );}for( int i = 0;i < n;i ++ ) {addedge( i, i + 1, 1 );addedge( i + 1, i, 0 );}memset( dis, 0x3f, sizeof( dis ) );dis[0] = 0; q.push( make_pair( 0, 0 ) );while( ! q.empty() ) {Pair now = q.top(); q.pop();int u = now.second;if( vis[u] or dis[u] ^ now.first ) continue;vis[u] = 1;for( auto nxt : G[u] ) {int v = nxt.first, w = nxt.second;if( dis[v] > dis[u] + w ) {dis[v] = dis[u] + w;q.push( make_pair( dis[v], v ) );}}}for( int i = 1;i <= n;i ++ )printf( "%d ", ( dis[i] - dis[i - 1] ) ^ 1 );return 0;
}