Codeforces Round #224 (Div. 2)(数学、dfs)

A.Ksenia and Pan Scales

好久没写代码了，写的太烂了。。。

#include<iostream>#include<string>#include<cstdio>using namespace std;int Abs(int x){    return x>0?x:(-x);}int main(){    //freopen("in.txt","r",stdin);    string a,b;    int numl=0,numr=0;    bool flag=false;    cin>>a>>b;    for(int i=0;i<a.size();i++)    {        if(!flag&&a[i]!='|')            numl++;        else if(a[i]=='|')        {            flag=true;        }        else numr++;    }    if(Abs(numr-numl)<=b.size())    {        int m=Abs(numr-numl);        int p=m+b.size();        if(p%2==0)        {            if(m==b.size())            {                if(numl>numr)cout<<a<<b;                else cout<<b<<a;            }            else            {                string tmp(b.begin(),b.begin()+p/2);                string tmp1(b.begin()+p/2,b.end());                if(numl<numr)cout<<tmp<<a<<tmp1<<endl;                else cout<<tmp1<<a<<tmp<<endl;            }        }        else cout<<"Impossible"<<endl;    }    else cout<<"Impossible"<<endl;    return 0;}

B. Number Busters

题意：求经过多少时间c<=a，c和a都有变换的方式。

思路：找循环节，c每秒都减1，a在b<x的时候减1,并且b = b + w - x,在b>=x的时候不变化，并且b = b - x..也就是a和c只有在b = b - x的时候差距才会减少1,所以b一共减了(c-a)次x, 设加了k次w - x.那么得到b最后的值为b - x * (c - a) + k * (w - x).然后b最后的值+x 必然不小于x(因为b >= x 才是执行b = b - x)..所以得到公式b - x * (c - a) + k *(w - x) + x >= x ==> k = (ceil)((b - x * (c - a))/(w - x)).

刚开始自己写了一个半暴力，挂了，果断超时，却只需要一个小小的公式，弱爆了。

#include<iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;long long a, b, w, x, c;long long solve() {if (c <= a) return 0;long long ans = (long long)ceil((x * (c - a) - b) * 1.0 / (w - x)) + (c - a);return ans;}int main() {cin>>a>>b>>w>>x>>c;cout<<solve()<<endl;return 0;}

C. Arithmetic Progression

给你n个数，让你添加一个数使其成为等差数列，输出所有符合要求的数。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;int n,a[maxn];int main(){    //freopen("in.txt","r",stdin);    cin>>n;    for(int i=1;i<=n;i++)        cin>>a[i];    if(n==1){cout<<-1<<endl;return 0;}    sort(a+1,a+n+1);    int num=0,x=a[2]-a[1],y=-1,numx=1,numy=0,biao;    for(int i=3;i<=n;i++)    {        int tmp=a[i]-a[i-1];        if(tmp==x)numx++;        else if(tmp!=x&&y==-1){y=tmp;biao=i-1;numy++;}        else if(tmp!=x&&tmp==y)numy++;        else if(tmp!=x&&tmp!=y) {cout<<0<<endl;return 0;}    }    if(numx>1&&numy>1){cout<<0;return 0;}    if(numx==1&&x>y)biao=1;    if((numx==1&&numy==1)||(numx==1&&numy>1&&x>y)||(numy==1&&x<y))    {        int p=min(x,y);        int q=max(x,y);        if(p*2!=q){cout<<0<<endl;return 0;}        cout<<1<<endl<<a[biao]+p<<endl;    }    else if(numy==0)    {        if(x==0){cout<<1<<endl<<a[1]<<endl;return 0;}        int tmp=a[2]-a[1];        if(n==2&&(tmp)%2==0)            cout<<3<<endl<<a[1]-tmp<<" "<<a[1]+tmp/2<<" "<<a[2]+tmp<<endl;        else cout<<2<<endl<<a[1]-tmp<<" "<<a[n]+tmp<<endl;    }    else cout<<0<<endl;    return 0;}

D. Ksenia and Pawns

思路：先对每个格子进行dfs记录下从当前格子直到遇到#可以走多少步，并举录下最大的步数的格子，然后验证最大的格子是否有相交的时候。

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int MAXN=2010;const int INF=1000000000;struct Node{    int x,y;} node[2*MAXN];char grid[MAXN][MAXN];int num[MAXN][MAXN];bool vis[MAXN][MAXN];int n,m,ansn,k,flag;void init(){    memset(vis,false,sizeof(vis));    memset(num,-1,sizeof(num));}int dfs(int x,int y){    int t;    if(num[x][y]!=-1)return num[x][y];    if(grid[x][y]=='^')    {        if(vis[x-1][y]) return INF;        vis[x-1][y]=1;        t=dfs(x-1,y);        vis[x-1][y]=0;    }    else if(grid[x][y]=='v')    {        if(vis[x+1][y]) return INF;        vis[x+1][y]=1;        t=dfs(x+1,y);        vis[x+1][y]=0;    }    else if(grid[x][y]=='<')    {        if(vis[x][y-1]) return INF;        vis[x][y-1]=1;        t=dfs(x,y-1);        vis[x][y-1]=0;    }    else if(grid[x][y]=='>')    {        if(vis[x][y+1]) return INF;        vis[x][y+1]=1;        t=dfs(x,y+1);        vis[x][y+1]=0;    }    else t=-1;    num[x][y]=t+1;    return num[x][y];}void dfs1(int x,int y){    if(grid[x][y]=='#')    {        if(k) flag=1;        return ;    }    vis[x][y]=1;    if(grid[x][y]=='^')    {        if(!vis[x-1][y]) dfs1(x-1,y);    }    else if(grid[x][y]=='v')    {        if(!vis[x+1][y]) dfs1(x+1,y);    }    else if(grid[x][y]=='<')    {        if(!vis[x][y-1]) dfs1(x,y-1);    }    else if(grid[x][y]=='>')    {        if(!vis[x][y+1]) dfs1(x,y+1);    }}void solve(){    int tmp,cnt=0,ans=-1;    for(int i=1; i<=n; i++)    {        for(int j=1; j<=m; j++)        {            if(grid[i][j]=='#')continue;            vis[i][j]=true;            tmp=dfs(i,j);            vis[i][j]=0;            if(tmp>ans)            {                ans=tmp;                cnt=1;                node[cnt].x=i,node[cnt].y=j;            }            else if(tmp==ans)            {                node[++cnt].x=i,node[cnt].y=j;            }            if(ans>=INF)            {                ansn=-1;                return;            }        }    }    if(ans==-1)    {        ansn=0;        return;    }    memset(vis,0,sizeof(vis));    k=0;    dfs1(node[1].x,node[1].y);    flag=0;    k=1;    for(int i=2; i<=cnt; i++)    {        dfs1(node[i].x,node[i].y);        if(flag) break ;    }    ansn=2*ans-1+flag;}int main(){    //freopen("in.txt","r",stdin);    cin>>n>>m;    for(int i=1; i<=n; i++)        cin>>grid[i]+1;    init();    solve();    cout<<ansn<<endl;    return 0;}