文章目錄

• A - Chinchirorin
• B - AtCoder Condominium
• C - Friends and Travel costs
• D - Pond
• E - White Pawn
• F - Weed

ABC203

A - Chinchirorin

三個條件if判

#include <cstdio> int main() {int a, b, c;scanf( "%d %d %d", &a, &b, &c );if( a == b ) return ! printf( "%d\n", c );if( a == c ) return ! printf( "%d\n", b );if( b == c ) return ! printf( "%d\n", a );printf( "0\n" );return 0; }

B - AtCoder Condominium

for循環

#include <cstdio> int main() {int n, k;scanf( "%d %d", &n, &k );int ans = 0;for( int i = 1;i <= n;i ++ )for( int j = 1;j <= k;j ++ )ans += i * 100 + j;printf( "%d\n", ans );return 0; }

C - Friends and Travel costs

sort后再for掃一遍

#include <cstdio> #include <algorithm> using namespace std; #define int long long #define maxn 200005 struct node {int pos, w; }p[maxn]; int n, k;bool cmp( node x, node y ) {return x.pos < y.pos; }signed main() {scanf( "%lld %lld", &n, &k );for( int i = 1;i <= n;i ++ )scanf( "%lld %lld", &p[i].pos, &p[i].w );sort( p + 1, p + n + 1, cmp );int ans = k, pos = 0;for( int i = 1;i <= n;i ++ )if( pos + ans >= p[i].pos ) {ans -= ( p[i].pos - pos );ans += p[i].w;pos = p[i].pos;}else break;printf( "%lld\n", pos + ans );return 0; }

D - Pond

非常樸素的扣出每一個正方形暴力做，$O(nm{k}^2)$

就是把每個數當成可能成為的答案做

既然每個數都來一次超時，那就二分

二分中位數，用二維前綴和做

設$f_{i,j}:$ $k\times k$正方形的右下角為$(i,j)$，該正方形中嚴格大于中位數的個數

設$su{m}_{i,j}:$ 第$j$列的$1\to i$中嚴格大于中位數的個數

中位數在第$?\frac{k^2}{2}?+1$位，那么嚴格大于的個數就為$?\frac{k^2}{2}?$

只需要判斷有沒有某個正方形中$f_{i,j}$比要求個數小的，這說明該正方形的中位數應該更小，那么二分值往下調；否則往上調

#include <cstdio> #include <iostream> using namespace std; #define int long long #define maxn 805 int n, k, t, ans = 1e18; int a[maxn][maxn], sum[maxn][maxn], f[maxn][maxn];int c( int x ) {if( x < 0 ) return 0;else return x; }bool check( int x ) {for( int i = 1;i <= n;i ++ )for( int j = 1;j <= n;j ++ )sum[i][j] = ( a[i][j] > x ), f[i][j] = 0;for( int i = 1;i <= n;i ++ )for( int j = 1;j <= n;j ++ )sum[j][i] += sum[j - 1][i];for( int i = 1;i <= n;i ++ )for( int j = 1;j <= n;j ++ ) {f[i][j] = f[i][j - 1] - ( sum[i][c(j - k)] - sum[c(i - k)][c(j - k)] ) + ( sum[i][j] - sum[c(i - k)][j] );if( i >= k && j >= k && f[i][j] <= t ) return 1;}return 0; }signed main() {scanf( "%lld %lld", &n, &k );for( int i = 1;i <= n;i ++ )for( int j = 1;j <= n;j ++ )scanf( "%lld", &a[i][j] );t = k * k / 2;int l = 0, r = 1e9;while( l <= r ) {int mid = ( l + r ) >> 1;if( check( mid ) ) ans = mid, r = mid - 1;else l = mid + 1;}printf( "%lld\n", ans );return 0; }

E - White Pawn

如果$i+1$行一個黑格子都沒有，那么$i+1$行的狀態與$i$是一樣的，$n$的級別一下子被壓成了$m$級別

每一層最多往外擴$2$，那么整體來說到達點數硬存下來也是可行的

設$S_i=\{whitecanreach(x,y)indept{h}_i\}$

• $(j?1\in S_{i?1}?j+1\in S_{i?1})?j\in /S_{i?1}?j\in S_i$

  因為寫法其實$S$是一個，如果$j$之前已經在里面了，就沒必要加了(雖然用的是自動去重set)

• $j?1\in /S_{i?1}?j+1\in /S_{i?1}?j\in S_{i?1}?j\in /S_i$

#include <set> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; set < int > s; vector < int > Old, New; vector < pair < int, int > > g; int n, m;int main() {scanf( "%d %d", &n, &m );for( int i = 0, x, y;i < m;i ++ ) {scanf( "%d %d", &x, &y );g.push_back( make_pair( x, y ) );}sort( g.begin(), g.end() );s.insert( n );for( int l = 0, r;l < m;l = r ) {for( r = l;r < m && g[l].first == g[r].first;r ++ );New.clear(), Old.clear();for( int i = l;i < r;i ++ ) {int k = g[i].second;if( ( s.find( k - 1 ) != s.end() || s.find( k + 1 ) != s.end() ) && ( s.find( k ) == s.end() ) ) New.push_back( k );if( ( s.find( k - 1 ) == s.end() && s.find( k + 1 ) == s.end() ) && ( s.find( k ) != s.end() ) ) Old.push_back( k );}for( int i = 0;i < New.size();i ++ ) s.insert( New[i] );for( int i = 0;i < Old.size();i ++ ) s.erase( Old[i] );}printf( "%d\n", s.size() );return 0; }

F - Weed

由于特殊操作$?\frac{H}{2}?$在，不難發現，操作次數是$\log N$級別的

將雜草高度排序，直接枚舉操作次數，記$f_{i,j}:$ 第$i$次操作到$j$雜草為止拔草的最大數目

$f_{i?1,k}+{\sum }_{a_k>?\frac{a_j}{2}?,k\le j}1\to f_{i,j}$

最后滾動掉$i$這一維即可

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; #define maxn 200005 int f[2][maxn]; int a[maxn]; int n, k;int main() {scanf( "%d %d", &n, &k );for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] );if( n == k ) return ! printf( "0 %d\n", n );sort( a + 1, a + n + 1 );int ans = 0;for( int i = 1;i <= 31;i ++ ) {int ip = 0, maxx = 0;for( int j = 1;j <= n;j ++ ) {while( ip < n && a[ip + 1] <= a[j] / 2 )maxx = max( maxx, f[i & 1 ^ 1][++ ip] );f[i & 1][j] = maxx + j - ip;ans = max( ans, f[i & 1][j] );}if( ans >= n - k ) return ! printf( "%d %d\n", i, n - ans );}return 0; }

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