AtCoder Beginner Contest 176总结

由于打球又鸽了一场，快开学了好好打球！！！（狗头还是补一补

A - Takoyaki

签到题

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std;
int n,x,t;
int main()
{cin>>n>>x>>t;cout<<(n+x-1)/x*t<<endl;return 0;
}

B - Multiple of 9

第一次打学校比赛的时候就有一个类似的题
不过这题直接把结论告诉了一个数是9的倍数充要条件是十进制下每一位加起来的和是9的倍数
直接让这题称为模拟题

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
string s;
int main()
{cin>>s;ll res=0;for(auto t:s) res+=t-'0';if(res%9==0) cout<<"Yes"<<endl;else cout<<"No"<<endl;return 0;
}

C - Step

记录last前面最大的数，然后更新答案即可

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=200010;
int a[N],n;
int main()
{cin>>n;int last=0;ll res=0;for(int i=1;i<=n;i++) cin>>a[i];for(int i=1;i<=n;i++){if(a[i]<last) res+=last-a[i];else last=a[i];}cout<<res<<endl;return 0;
}

D - Wizard in Maze

双端队列bfs，使用魔法代价为1其他代价为0，更新即可。

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1010;
pii start,ed;
int n,m;
char g[N][N];
int dist[N][N];
bool st[N][N];
void bfs()
{memset(dist,0x3f,sizeof dist);deque<pii> q;dist[start.x][start.y]=0;q.push_front(start);int dx[]={0,0,1,-1},dy[]={1,-1,0,0};while(q.size()){auto t=q.front();q.pop_front();if(st[t.x][t.y]) continue;st[t.x][t.y]=1;int d=dist[t.x][t.y];int a,b;for(int i=0;i<4;i++){a=t.x+dx[i],b=t.y+dy[i];if(a<1||b<1||a>n||b>m||g[a][b]=='#'||dist[a][b]<=d) continue;dist[a][b]=d;q.push_front({a,b});}for(int i=t.x-2;i<=t.x+2;i++)   {if(i<1||i>n) continue;for(int j=t.y-2;j<=t.y+2;j++){if(j<1||j>m||g[i][j]=='#'||dist[i][j]<=d+1) continue;dist[i][j]=d+1;q.push_back({i,j});}}}
}
int main()
{cin>>n>>m;cin>>start.x>>start.y;cin>>ed.x>>ed.y;for(int i=1;i<=n;i++) cin>>g[i]+1;bfs();if(dist[ed.x][ed.y]==0x3f3f3f3f) cout<<-1<<endl;else cout<<dist[ed.x][ed.y]<<endl;return 0;
}

E - Bomber

直接枚举懒得写题解了

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=300010;
int n,m;
int k;
map<pii,int> mp;
int col[N],row[N];
int maxr,maxc;
vector<int> a,b;
int main()
{cin>>n>>m>>k;while(k--){int a,b;cin>>a>>b;mp[{a,b}]=1;row[a]++;col[b]++;maxr=max(maxr,row[a]);maxc=max(maxc,col[b]);}for(int i=1;i<=n;i++) if(row[i]==maxr) a.push_back(i);for(int j=1;j<=m;j++)if(col[j]==maxc) b.push_back(j);for(int i=0;i<a.size();i++) for(int j=0;j<b.size();j++)if(mp[{a[i],b[j]}]==0) {cout<<maxr+maxc<<endl;return 0;}cout<<maxr+maxc-1<<endl;return 0;
}