Codeforces Round #684 (Div. 2)

A - Buy the String

要么全变成1要么全变成0要么一个都不改变，三种情况取最小。

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int main()
{IO;int T=1;cin>>T;while(T--){int n;cin>>n;int a,b,c;cin>>a>>b>>c;string s;cin>>s;int res1=0;for(int i=0;i<n;i++){if(s[i]=='0') res1+=a;else res1+=b;}int res2=0;for(int i=0;i<n;i++){if(s[i]=='0') res2+=b+c;else res2+=b;}int res3=0;for(int i=0;i<n;i++){if(s[i]=='0') res3+=a;else res3+=a+c;}cout<<min(res1,min(res2,res3))<<'\n';}return 0;
}

B - Sum of Medians

数学题，取后面几个中位数即可
任意考虑一个不降序列a1a2a3…ana_1 \ a_2 \ a_3\dots a_na1 a2 a3…an，按照题目意思中位数为a⌊1+n2⌋a_{\lfloor \frac{1+n}{2}\rfloor}a⌊21+n⌋，我们那么中位数后面的数的个数为n−⌊1+n2⌋n-\lfloor \frac{1+n}{2}\rfloorn−⌊21+n⌋意味着如果你选择一个数作为中位数那么一定要有n−⌊1+n2⌋n-\lfloor \frac{1+n}{2}\rfloorn−⌊21+n⌋不小于它的数。

那么本题只需要每次选择第n−⌊1+n2⌋+1n-\lfloor \frac{1+n}{2}\rfloor+1n−⌊21+n⌋+1大的数作为每个数组的中位数即可。

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1000010;
int n,k;
ll a[N];
int main()
{IO;int T=1;cin>>T;while(T--){cin>>n>>k;for(int i=1;i<=n*k;i++) {a[i]=0;cin>>a[i];}sort(a+1,a+1+n*k);int m=(n+1)/2;int cnt=n-m;ll res=0;for(int i=n*k,j=1;i>=1,j<=k;j++,i-=cnt+1)res+=a[i-cnt];cout<<res<<'\n';}return 0;
}

C1 - Binary Table (Easy Version)

[1111]→[1000]→[1100]→[1110]→[0000]\begin{bmatrix}1 &1\\ 1&1\end{bmatrix}\to \begin{bmatrix}1 &0\\ 0&0\end{bmatrix}\to \begin{bmatrix}1 &1\\ 0&0\end{bmatrix} \to \begin{bmatrix}1 &1\\ 1&0\end{bmatrix} \to \begin{bmatrix}0 &0\\ 0&0\end{bmatrix}[1111]→[1000]→[1010]→[1110]→[0000]
每次考虑变换4个格子，每4个格子最多变换4次，我们会把4个数全部变成0，如果行列都是偶数，那么保证mn次操作内绝对能够全变成0
由于行列奇偶行可能导致出现某些元素重复变换导致nm次过不了（这也是更难的版本

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=110;
int n,m;
char tmp[N][N];
int g[N][N];
pii a[4];
bool st[4];
vector<pii> ans;
void work(pii a[])
{memset(st,0,sizeof st);int cnt=0;for(int i=0;i<4;i++){int x=a[i].first,y=a[i].second;cnt+=g[x][y];}if(cnt==1){int now=0;for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(now<2&&!g[x][y]){st[i]=1;ans.push_back({x,y});g[x][y]=1;now++;}}for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!st[i]&&g[x][y]){ans.push_back({x,y});g[x][y]=0;break;}}work(a);}else if(cnt==2){for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!g[x][y]){st[i]=1;ans.push_back({x,y});g[x][y]=1;}}for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!st[i]&&g[x][y]){ans.push_back({x,y});g[x][y]=0;break;}}work(a);}else if(cnt==3){for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(g[x][y]){ans.push_back({x,y});g[x][y]=0;}}}else if(cnt==4){for(int i=0;i<3;i++) {int x=a[i].first,y=a[i].second;if(g[x][y]){ans.push_back({x,y});g[x][y]=0;}}work(a);}else return;
}
int main()
{IO;int T=1;cin>>T;while(T--){cin>>n>>m;for(int i=1;i<=n;i++) cin>>tmp[i]+1;for(int i=1;i<=n;i++)   for(int j=1;j<=m;j++)   g[i][j]=tmp[i][j]-'0';for(int i=1;i<=n;i+=2)for(int j=1;j<=m;j+=2){if(i+1<=n&&j+1<=m)a[0]={i,j},a[1]={i+1,j},a[2]={i,j+1},a[3]={i+1,j+1};else if(i+1>n&&j+1<=m)a[0]={i,j},a[1]={i-1,j},a[2]={i,j+1},a[3]={i-1,j+1};else if(i+1<=n&&j+1>m)a[0]={i,j},a[1]={i+1,j},a[2]={i,j-1},a[3]={i+1,j-1};elsea[0]={i,j},a[1]={i-1,j},a[2]={i,j-1},a[3]={i-1,j-1};work(a);}cout<<ans.size()/3<<'\n';int cnt=0;for(auto t:ans){cnt++;cout<<t.first<<' '<<t.second<<' ';if(cnt%3==0) cout<<'\n';}ans.clear();}return 0;
}

C2 - Binary Table (Hard Version)

对于行列数是奇数可能导致nm操作不能搞的情况：其实我们只要首先把一行全部变成0（行数是奇数）或者一列全部变成0（列数是奇数），我们就能把操作次数控制在n2+1+m2+1+(m−1)×(n−1)\frac{n}{2}+1+\frac{m}{2}+1+(m-1)×(n-1)2n+1+2m+1+(m−1)×(n−1)次以内

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=110;
int n,m;
char tmp[N][N];
int g[N][N];
pii a[4];
bool st[4];
vector<pii> ans;
void work(pii a[])
{memset(st,0,sizeof st);int cnt=0;for(int i=0;i<4;i++){int x=a[i].first,y=a[i].second;cnt+=g[x][y];}if(cnt==1){int now=0;for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(now<2&&!g[x][y]){st[i]=1;ans.push_back({x,y});g[x][y]=1;now++;}}for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!st[i]&&g[x][y]){ans.push_back({x,y});g[x][y]=0;break;}}work(a);}else if(cnt==2){for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!g[x][y]){st[i]=1;ans.push_back({x,y});g[x][y]=1;}}for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(!st[i]&&g[x][y]){ans.push_back({x,y});g[x][y]=0;break;}}work(a);}else if(cnt==3){for(int i=0;i<4;i++) {int x=a[i].first,y=a[i].second;if(g[x][y]){ans.push_back({x,y});g[x][y]=0;}}}else if(cnt==4){for(int i=0;i<3;i++) {int x=a[i].first,y=a[i].second;if(g[x][y]){ans.push_back({x,y});g[x][y]=0;}}work(a);}else return;
}
void print()
{for(int i=1;i<=n;i++){for(int j=1;j<=m;j++) cout<<g[i][j]<<' ';cout<<endl;}
}
int main()
{IO;int T=1;cin>>T;while(T--){cin>>n>>m;for(int i=1;i<=n;i++) cin>>tmp[i]+1;for(int i=1;i<=n;i++)   for(int j=1;j<=m;j++)   g[i][j]=tmp[i][j]-'0';if(n%2==0&&m%2==0){for(int i=1;i<=n;i+=2)for(int j=1;j<=m;j+=2){a[0]={i,j},a[1]={i+1,j},a[2]={i,j+1},a[3]={i+1,j+1};work(a);}}else if(n%2==0&&m%2!=0){for(int i=1;i<=n;i+=2){if(g[i][1]==0&&g[i+1][1]==0) continue;else if(g[i][1]==1&&g[i+1][1]==0){ans.push_back({i,1}); g[i][1]=1-g[i][1];ans.push_back({i,2}); g[i][2]=1-g[i][2];ans.push_back({i+1,2}); g[i+1][2]=1-g[i+1][2];}else if(g[i][1]==0&&g[i+1][1]==1){ans.push_back({i+1,1}); g[i+1][1]=1-g[i+1][1];ans.push_back({i,2}); g[i][2]=1-g[i][2];ans.push_back({i+1,2}); g[i+1][2]=1-g[i+1][2];}else {ans.push_back({i,1}); g[i][1]=1-g[i][1];ans.push_back({i+1,1}); g[i+1][1]=1-g[i+1][1];ans.push_back({i+1,2}); g[i+1][2]=1-g[i+1][2];}}for(int i=1;i<=n;i+=2)for(int j=2;j<=m;j+=2){a[0]={i,j},a[1]={i+1,j},a[2]={i,j+1},a[3]={i+1,j+1};work(a);}}else if(n%2!=0&&m%2==0){for(int j=1;j<=m;j+=2){if(g[1][j]==0&&g[1][j+1]==0) continue;else if(g[1][j]==1&&g[1][j+1]==0){ans.push_back({1,j});g[1][j]=1-g[1][j];ans.push_back({2,j});g[2][j]=1-g[2][j];ans.push_back({2,j+1});g[2][j+1]=1-g[2][j+1];}else if(g[1][j]==0&&g[1][j+1]==1){ans.push_back({1,j+1});g[1][j+1]=1-g[1][j+1];ans.push_back({2,j});g[2][j]=1-g[2][j];ans.push_back({2,j+1});g[2][j+1]=1-g[2][j+1];}else {ans.push_back({1,j});g[1][j]=1-g[1][j];ans.push_back({1,j+1});g[1][j+1]=1-g[1][j+1];ans.push_back({2,j});g[2][j]=1-g[2][j];}}for(int i=2;i<=n;i+=2)for(int j=1;j<=m;j+=2){a[0]={i,j},a[1]={i+1,j},a[2]={i,j+1},a[3]={i+1,j+1};work(a);}//print();}else{for(int i=1;i<=n;i+=2){int x=i+1;if(x>n) x-=2;if(g[i][1]==0&&g[x][1]==0) continue;else if(g[i][1]==1&&g[x][1]==0){ans.push_back({i,1}); g[i][1]=1-g[i][1];ans.push_back({i,2}); g[i][2]=1-g[i][2];ans.push_back({x,2}); g[x][2]=1-g[x][2];}else if(g[i][1]==0&&g[x][1]==1){ans.push_back({x,1}); g[x][1]=1-g[x][1];ans.push_back({i,2}); g[i][2]=1-g[i][2];ans.push_back({x,2}); g[x][2]=1-g[x][2];}else {ans.push_back({x,1}); g[x][1]=1-g[x][1];ans.push_back({i,1}); g[i][1]=1-g[i][1];ans.push_back({x,2}); g[x][2]=1-g[x][2];}}for(int j=2;j<=m;j+=2){int y=j+1;if(y>m) y-=2;if(g[1][j]==0&&g[1][y]==0) continue;else if(g[1][j]==1&&g[1][y]==0){ans.push_back({1,j});g[1][j]=1-g[1][j];ans.push_back({2,j});g[2][j]=1-g[2][j];ans.push_back({2,y});g[2][y]=1-g[2][y];}else if(g[1][j]==0&&g[1][y]==1){ans.push_back({1,y});g[1][y]=1-g[1][y];ans.push_back({2,j});g[2][j]=1-g[2][j];ans.push_back({2,y});g[2][y]=1-g[2][y];}else {ans.push_back({1,j});g[1][j]=1-g[1][j];ans.push_back({2,j});g[2][j]=1-g[2][j];ans.push_back({1,y});g[1][y]=1-g[1][y];}}//print();for(int i=2;i<=n;i+=2)for(int j=2;j<=m;j+=2){a[0]={i,j},a[1]={i+1,j},a[2]={i,j+1},a[3]={i+1,j+1};work(a);}}cout<<ans.size()/3<<'\n';int cnt=0;for(auto t:ans){cnt++;cout<<t.first<<' '<<t.second<<' ';if(cnt%3==0) cout<<'\n';}ans.clear();}return 0;
}

这模拟就离谱好吧~~

写了4题太晚了（希望不被fst，就没有在往后看

E. Greedy Shopping

今天补了一下E. Greedy Shopping
要加油哦~