AtCoder Beginner Contest 182 题解

传送门：
https://atcoder.jp/contests/abc182

atcoder 好久没补了Orz，今天数据结构、马原课写了一下（

A

#include<bits/stdc++.h>
using namespace std;

int main(){
	int a, b; cin>>a>>b;
	cout<<2*a+100-b;	
	return 0;
}

B

#include<bits/stdc++.h>
using namespace std;

const int N=105;

int w[N], n;

int main(){
	cin>>n;
	for(int i=1; i<=n; i++) cin>>w[i];
	
	int len=0, res;
	for(int i=2; i<=1000; i++){
		int cnt=0;
		for(int j=1; j<=n; j++) if(w[j]%i==0) cnt++;
		if(cnt>len){
			len=cnt;
			res=i;
		}
	}
	cout<<res;
	
	return 0;
}

C

我去年十一月现场打过这个比赛，因此第一份是去年的(码风毒瘤)。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath> 
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<list>
#include<cstdlib>
#include<set>
#include<map>
using namespace std;
typedef unsigned long long ll;

int main(){
	string s;
	while(cin>>s){
		int rec1=0,rec2=0,rec3=0;
		int sum=0;
		for(int i=0;i<s.length();i++){
			int cnt=s[i]-'0';
			if(cnt %3 ==1) rec1++;
			else if(cnt %3 ==2) rec2++;
			else rec3++;
			sum+=cnt;
		}

		int len=s.length();
		//cout<<sum;
		if(sum%3==0)cout<<0;
		else{
			if(len<=2 && !rec3) cout<<-1;
			else if(!rec2){
				if(sum%3==1) cout<<1;
				else cout<<2;
			}
			else if(!rec1){
				if(sum%3==2) cout<<1;
				else cout<<2;
			}
			else{
				cout<<1;
			}
		}
	}
	
	
	return 0;
}

#include<bits/stdc++.h>
using namespace std;

int lowbit(int x){
	return x&-x;
}

int cal(int x){
	int res=0;
	while(x) x-=lowbit(x), res++;
	return res;	
}

int main(){
	string s; cin>>s;
	int n=s.size();
	
	int res=0;
	for(int i=0; i<(1<<n); i++){
		int r=0;
		for(int j=0; j<n; j++) if(i>>j&1) r+=s[j]-'0';
		if(r%3==0) res=max(res, cal(i));
	}
	
	if(!res) puts("-1");
	else cout<<n-res<<endl;
	
	return 0;
}

D

直接枚举前面走过的完整行对应的贡献 + 当前行最大前缀和（这个可以一边扫一边维护）。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

#define int ll

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=2e5+5;

int n, w[N];
int f[N], mx[N];

signed main(){
	read(n);
	rep(i,1,n) read(w[i]), w[i]+=w[i-1];
	
	int res=0;
	
	rep(i,1,n){
		if(i!=1) mx[i]=max(w[i], mx[i-1]);
		else mx[i]=w[i];
		
		int t=f[i-1]+mx[i];
		res=max(res, t);
		f[i]=w[i]+f[i-1];
	}
	cout<<res<<endl;
    return 0;
}

E

从四个方向更新可以照到的地方。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb push_back
#define eb emplace_back
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1550;

int n, m;
int w[N][N];
int g[4][N][N];

int main(){
	read(n), read(m);
	int u, v; read(u), read(v);
	rep(i,1,u){
		int x, y; read(x), read(y);
		w[x][y]=1;
	}
	rep(i,1,v){
		int x, y; read(x), read(y);
		w[x][y]=-1;
	}
	
	rep(i,0,3) memcpy(g[i], w, sizeof g[i]);

	rep(i,1,n) rep(j,1,m){
		if(w[i][j]==-1) g[0][i][j]=0;
		else if(g[0][i][j-1]==1 || w[i][j]==1) g[0][i][j]=1;
		else if(w[i][j-1]==-1) g[0][i][j]=0;
	}
	rep(i,1,n) dwn(j,m,1){
		if(w[i][j]==-1) g[1][i][j]=0;
		else if(g[1][i][j+1]==1 || w[i][j]==1) g[1][i][j]=1;
		else if(w[i][j+1]==-1) g[1][i][j]=0;
	}
	cerr<<g[1][3][1]<<endl;
	rep(j,1,m) rep(i,1,n){
		if(w[i][j]==-1) g[2][i][j]=0;
		else if(g[2][i-1][j]==1 || w[i][j]==1) g[2][i][j]=1;
		else if(w[i-1][j]==-1) g[2][i][j]=0;
	}
	rep(j,1,m) dwn(i,n,1){
		if(w[i][j]==-1) g[3][i][j]=0;
		else if(g[3][i+1][j]==1 || w[i][j]==1) g[3][i][j]=1;
		else if(w[i+1][j]==-1) g[3][i][j]=0;
	}
	
	int res=0;
	rep(i,1,n){
		rep(j,1,m){
			int t=0;
			rep(x,0,3) t|=g[x][i][j]; 
			res+=t;
		} 
	}
	cout<<res<<endl;
	
    return 0;
}

标签：AtCoder,ch,int,题解,long,182,using,include,define
来源： https://www.cnblogs.com/Tenshi/p/15240287.html