AtCoder Beginner Contest 420 (abc420) 赛后复盘
我再也不在赛时看瓜条了。。。
A&B
简单题，直接放代码。
A
#include<bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 200050
int x,y;
void work() {
cin>>x>>y;
cout<<(x-1+y)%12+1<<'\n';
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
// ios::sync_with_stdio(false);
// cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
B
#include<bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 200050
int n,m;
string s[maxn];
int pt[maxn];
void work() {
cin>>n>>m;
For(i,1,n) cin>>s[i],s[i]=" "+s[i];
For(i,1,m) {
int x=0,y=0;
For(j,1,n) if(s[j][i]=='0') x++;else y++;
if(x!=0&&y!=0) {
if(x<y) {For(j,1,n) if(s[j][i]=='0') pt[j]++;}
else {For(j,1,n) if(s[j][i]=='1') pt[j]++;}
}
}
int res=0;
For(i,1,n) res=max(res,pt[i]);
For(i,1,n) if(pt[i]==res) cout<<i<<' ';
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
// ios::sync_with_stdio(false);
// cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
C
注意到每次修改只会对一位产生影响，所以只用先算出答案，每次操作对当前位置更新计算就好了。
点击查看代码
#include<bits/stdc++.h>
#define int ll
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 200050
int n,q;
int a[maxn],b[maxn];
void work() {
in2(n,q);
For(i,1,n) in1(a[i]);
For(i,1,n) in1(b[i]);
int res=0;
For(i,1,n) res+=min(a[i],b[i]);
while(q--) {
char opt=getchar();
while(opt!='A'&&opt!='B') opt=getchar();
int x,y; in2(x,y);
res-=min(a[x],b[x]);
if(opt=='A') {
a[x]=y;
}
if(opt=='B') {
b[x]=y;
}
res+=min(a[x],b[x]);
cout<<res<<'\n';
}
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
// ios::sync_with_stdio(false);
// cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
D
注意到门开关只有两种状态，故 bfs 的时候再多记录一维表示当前状态即可。
点击查看代码
#include<bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 505
int vis[maxn][maxn][2];
int h,w;
char ch[maxn][maxn];
int sx,sy;
int dx[5]={0,1,-1,0,0},dy[5]={0,0,0,1,-1};
struct node {
int a,b,c,d;
};
void work() {
cin>>h>>w;
For(i,1,h) For(j,1,w) {
cin>>ch[i][j];
if(ch[i][j]=='S') sx=i,sy=j;
}
queue<node> q; q.push({sx,sy,0,0});
vis[sx][sy][0]=1;
while(!q.empty()) {
auto [x,y,flg,stp]=q.front(); q.pop();
if(ch[x][y]=='G') {
return cout<<stp,void();
}
For(i,1,4) {
int X=x+dx[i],Y=y+dy[i];
int f=flg^(ch[X][Y]=='?');
if(X>=1&&X<=h&&Y>=1&&Y<=w&&!vis[X][Y][f]&&ch[X][Y]!='#'&&!(((ch[X][Y]=='o')&&(flg==1))||((ch[X][Y]=='x')&&(flg==0)))) {
vis[X][Y][f]=1;
q.push({X,Y,f,stp+1});
}
}
}
cout<<-1<<'\n';
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
E
注意到一个点能到达的地方可以用并查集来维护，那么同时维护该并查集的黑点个数即可。对于答案即判断该集合黑点数是否为 \(0\) 即可。
点击查看代码
#include<bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 200050
int n,q;
int fa[maxn],siz[maxn],col[maxn];
int fnd(int x) {
return x==fa[x]?fa[x]:fa[x]=fnd(fa[x]);
}
void work() {
in2(n,q);
For(i,1,n) fa[i]=i;
while(q--) {
int opt,x,y;
in2(opt,x);
if(opt==1) {
in1(y);
y=fnd(y);
x=fnd(x);
if(x==y) continue;
fa[y]=x;
siz[x]+=siz[y];
} else if(opt==2){
if(col[x]) siz[fnd(x)]--,col[x]=0;
else siz[fnd(x)]++,col[x]=1;
} else {
cout<<(siz[fnd(x)]>0?"Yes":"No")<<'\n';
}
}
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
// ios::sync_with_stdio(false);
// cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
G
我们设 \(\sqrt{n^2+n+x}=(n+k)\)，那么有 \(n^2+n+x=n^2+2nk+k^2\)，即 \(x=k^2+n(2k-1)\)，因为 \(-10^{14}\le x \le 10^{14}\)，所以 \(|k|\le 10^7\)，那么枚举 \(k\) 即可。
由于我的代码实现有一些问题，每个数会被多算一遍，原因未知，因此需要 unique 一下。
点击查看代码
#include<bits/stdc++.h>
#define int i128
#define pii pair<int,int>
#define pll pair<long long,long long>
#define ll long long
#define i128 __int128
#define mem(a,b) memset((a),(b),sizeof(a))
#define m0(a) memset((a),0,sizeof(a))
#define m1(a) memset(a,-1,sizeof(a))
#define lb(x) ((x)&-(x))
#define lc(x) ((x)<<1)
#define rc(x) (((x)<<1)|1)
#define pb(G,x) (G).push_back((x))
#define For(a,b,c) for(int a=(b);a<=(c);a++)
#define Rep(a,b,c) for(int a=(b);a>=(c);a--)
#define in1(a) a=read()
#define in2(a,b) a=read(), b=read()
#define in3(a,b,c) a=read(), b=read(), c=read()
#define in4(a,b,c,d) a=read(), b=read(), c=read(), d=read()
#define fst first
#define scd second
#define dbg puts("IAKIOI")
using namespace std;
int read() {
int x=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar()) f=(c=='-'?-1:1);
for(;c<='9'&&c>='0';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
return x*f;
}
void write(int x) { if(x<0) {putchar('-'); write(-x); return ;}if(x>=10) write(x/10); putchar('0'+x%10); }
const int mod = 998244353;
int qpo(int a,int b) {int res=1; for(;b;b>>=1,a=(a*a)%mod) if(b&1) res=res*a%mod; return res; }
int inv(int a) {return qpo(a,mod-2); }
#define maxn 200050
int x;
vector<pii> a;
vector<int> ans;
bool flg=0;
void work() {
in1(x);
ans.push_back(x-1);
ans.push_back(-x);
for(int i=max(min(x-1,-x),(i128)-1e7-100);i<=min(max(x-1,-x),(i128)1e7+100);i++) {
if(i!=0&&(x-(i*i))%(i+i-1)==0) {
int n=(x-(i*i))/(i+i-1);
ans.push_back(n);
}
}
sort(ans.begin(),ans.end());
int len=unique(ans.begin(),ans.end())-ans.begin();
cout<<(ll)len<<'\n';
For(i,0,len-1) cout<<(ll)ans[i]<<' ';
}
signed main() {
// freopen("data.in","r",stdin);
// freopen("myans.out","w",stdout);
// ios::sync_with_stdio(false);
// cin.tie(0); cout.tie(0);
double stt=clock();
int _=1;
// _=read();
// cin>>_;
For(i,1,_) {
work();
}
cerr<<"\nTotal Time is:"<<(clock()-stt)*1.0/1000<<" second(s)."<<'\n';
return 0;
}
本文来自博客园，作者：coding_goat_qwq，转载请注明原文链接：https://www.cnblogs.com/CodingGoat/p/19055996
浙公网安备 33010602011771号