codeforces Intel Code Challenge Final Round (div.1 + div.2 combined)

比赛水掉3题rk559 rating+115

赛后切掉C

n年没打cf了终于又重新变蓝了，果然太弱。。。

1.A题 Checking the Calendar

　　给定两个星期几，问是否可能分别是两个月的第一天。

　　水题暴力枚举月份

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+'0';tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn (100050)
int n,m,k;
int d[4][2]={{1,1},{-1,1},{-1,-1},{1,-1}
};
struct node
{int x,y;ll tm;bool operator<(const node&rhs)const{if(x!=rhs.x)return x<rhs.x;else if(y!=rhs.y)return y<rhs.y;else return tm<rhs.tm;}
}p[maxn];
int nume;
node mp[4][maxn*20];
void read()
{cin>>n>>m>>k;FOR(i,1,k) {cin>>p[i].x>>p[i].y;}
}
void change(int &x,int &y,int &dir,int& dd)
{int nx,ny;if(dir==0){dd=min(n-x,m-y);nx=dd+x,ny=dd+y;if(nx==n&&ny==m) ;else if(nx==n) dir=1;else dir=3; }else if(dir==1){dd=min(x,m-y);nx=x-dd,ny=dd+y;if(nx==0&&ny==m) ;else if(nx==0) dir=0;else dir=2;}else if(dir==2){dd=min(x,y);nx=x-dd,ny=y-dd;if(nx==0&&ny==0) ;else if(nx==0) dir=3;else dir=1;}else{dd=min(n-x,y);nx=x+dd,ny=y-dd;if(nx==n&&ny==0) ;else if(nx==n) dir=2;else dir=0;}x=nx;y=ny;
}
void preprocess()
{int x=0,y=0,dir=0;ll sum=0;int cnt=0;while(cnt<=4*(m+n)){node tmp;tmp.x=x;tmp.y=y;tmp.tm=sum;int dd=0;mp[dir][nume++]=tmp;change(x,y,dir,dd);sum+=dd;cnt++;// ptl2;
    }FOR(j,0,3)sort(mp[j],mp[j]+nume);
}
void solve()
{FOR(i,1,k){ll ans=LLINF;FOR(j,0,3){int x=p[i].x,y=p[i].y,dir=j,dd=0;change(x,y,dir,dd);node tmp;tmp.x=x;tmp.y=y;tmp.tm=0;if(j==0) dir=2;else if(j==2) dir=0;else if(j==1) dir=3;else dir=1;FOR(t,0,3){int p=lower_bound(mp[t],mp[t]+nume,tmp)-mp[t];if(p>=0&&p<nume&&mp[t][p].x==x&&mp[t][p].y==y){ans=min(ans,mp[t][p].tm+dd);}}}printf("%I64d\n",ans==LLINF?-1:ans);}
}
int main()
{{read();preprocess();solve();}return 0;
}

View Code

2.B题 Batch Sort

　　给定矩阵要求最后每一行都排好序，支持操作a.每一行最多选两个交换一次 b.最后最多选两列整体交换一次。

　　似乎有很多人在这题上翻船了，估计是想复杂了，因为规模太小暴力枚举b，然后看看a行不行就行了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+'0';tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn 30
int T,N,M,mat[maxn][maxn],tmp[maxn][maxn],a[maxn];
void read()
{cin>>N>>M;FOR(i,1,N)FOR(j,1,M) cin>>mat[i][j];
}
void preprocess()
{}
int work(int i,int j)
{int flg=1;FOR(r,1,N){swap(tmp[r][i],tmp[r][j]);}FOR(r,1,N){FOR(c,1,M) a[c]=tmp[r][c];sort(a+1,a+1+M);int cnt=0;FOR(c,1,M) if(a[c]!=tmp[r][c]) cnt++;if(cnt>2) flg=0;}return flg;
}
void solve()
{int flg=0;FOR(i,1,M){FOR(j,i,M){memcpy(tmp,mat,sizeof(tmp));flg|=work(i,j);}}if(flg) puts("YES");else puts("NO");
}
int main()
{{read();solve();}return 0;
}

View Code

3.C题 Ray tracing

　　给定一个网格10w*10w，从(0,0)出发光线会反射，询问10w次某一个点能否到达。

　　比赛的时候是想的暴力模拟，记录所有经过的点，mle。于是尝试只记录边界点，tle。

　　正确的思路是暴力只模拟墙上落点，预处理出来。然后每个询问必然光线来自于墙上某个点,于是二分查找有没有这种点，注意用map记录落点会mle。

　　比赛的时候确实有过这种落点思路，但是没想到每个询问必然光线来自于墙上某个点，遗憾。。。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+'0';tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn (100050)
int n,m,k;
int d[4][2]={{1,1},{-1,1},{-1,-1},{1,-1}
};
struct node
{int x,y;ll tm;bool operator<(const node&rhs)const{if(x!=rhs.x)return x<rhs.x;else if(y!=rhs.y)return y<rhs.y;else return tm<rhs.tm;}
}p[maxn];
int nume;
node mp[4][maxn*20];
void read()
{cin>>n>>m>>k;FOR(i,1,k) {cin>>p[i].x>>p[i].y;}
}
void change(int &x,int &y,int &dir,int& dd)
{int nx,ny;if(dir==0){dd=min(n-x,m-y);nx=dd+x,ny=dd+y;if(nx==n&&ny==m) ;else if(nx==n) dir=1;else dir=3; }else if(dir==1){dd=min(x,m-y);nx=x-dd,ny=dd+y;if(nx==0&&ny==m) ;else if(nx==0) dir=0;else dir=2;}else if(dir==2){dd=min(x,y);nx=x-dd,ny=y-dd;if(nx==0&&ny==0) ;else if(nx==0) dir=3;else dir=1;}else{dd=min(n-x,y);nx=x+dd,ny=y-dd;if(nx==n&&ny==0) ;else if(nx==n) dir=2;else dir=0;}x=nx;y=ny;
}
void preprocess()
{int x=0,y=0,dir=0;ll sum=0;int cnt=0;while(cnt<=4*(m+n)){node tmp;tmp.x=x;tmp.y=y;tmp.tm=sum;int dd=0;mp[dir][nume++]=tmp;change(x,y,dir,dd);sum+=dd;cnt++;// ptl2;
    }FOR(j,0,3)sort(mp[j],mp[j]+nume);
}
void solve()
{FOR(i,1,k){ll ans=LLINF;FOR(j,0,3){int x=p[i].x,y=p[i].y,dir=j,dd=0;change(x,y,dir,dd);node tmp;tmp.x=x;tmp.y=y;tmp.tm=0;if(j==0) dir=2;else if(j==2) dir=0;else if(j==1) dir=3;else dir=1;FOR(t,0,3){int p=lower_bound(mp[t],mp[t]+nume,tmp)-mp[t];if(p>=0&&p<nume&&mp[t][p].x==x&&mp[t][p].y==y){ans=min(ans,mp[t][p].tm+dd);}}}printf("%I64d\n",ans==LLINF?-1:ans);}
}
int main()
{{read();preprocess();solve();}return 0;
}

View Code

4.D题 Dense subsequence

　　给定一个字母序列，要求选定子序列使得原来序列每m个连续子序列中至少包含一个选定子序列的元素。要求字典序最小。

　　从小到大枚举字母。如果尽量少的选当前字母能够达成条件，退出；否则删除这轮选的，全选序列中该字母。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<string>
#include<cctype>
#include<stack>
#include<queue>
#include<set>
#include<sstream>
#include<map>
using namespace std;
#define FORD(i,k,n) for(int i=n;i>=k;i--)
#define FOR(i,k,n) for(int i=k;i<=n;i++)
#define CLR(a,b) memset(a,b,sizeof(a));
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define LL long long
#define ull unsigned long long
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define ptb(b,a){int tmp=a;string s;do{s+=tmp%2+'0';tmp/=2;}while(tmp);reverse(s.begin(),s.end());cout<<"bin "<<b<<"="<<s<<endl;}
#define pta(i,a,f,b) {FOR(i,f,b) cout<<a[i]<<" "; printf("\n");}
#define pt(a,b) cout<<a<<"="<<b<<endl
#define pt1(a) cout<<a<<endl
#define pt2(a,b) cout<<a<<" "<<b<<endl
#define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define ptl1 cout<<"-------------"<<endl
#define ptl2 cout<<"~~~~~~~~~~~~~~~~"<<endl
#define fp   freopen("in.txt","r",stdin)
#define maxn 200000
int T,N,m,flg;
string  ans;
char s[maxn];
int last;
int vis[maxn];
void read()
{scanf("%d %s",&m,s+1);
}
void preprocess()
{
}
void solve()
{int len=strlen(s+1);FOR(i,'a','z'){last=0;int l=1,r=m;FOR(k,1,m) {if((last==0&&s[k]==i)||(s[k]<=s[last])||vis[k])last=k;}string tmp;if(last&&s[last]==i)tmp+=i;// pt1(last);while(r<len&&last){l++;r++;if(last>=l&&last<=r) continue;else{last=0;FOR(k,l,r){if((last==0&&s[k]==i)||(s[k]<=s[last])||vis[k]) last=k;}if(last&&s[last]==i) {tmp+=i;}}// pt4(last,l,r,ans);
        }if(last){ans+=tmp;break;}else{FOR(k,1,len){if(s[k]==i) {vis[k]=1;ans+=i;}}}}cout<<ans<<endl;
}
int main()
{{read();solve();}return 0;
}

View Code

转载于:https://www.cnblogs.com/diang/p/5944230.html

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