Codeforces Round #354 (Div. 2)

比赛链接：

http://codeforces.com/contest/676

A. Nicholas and Permutation

分析：

水题，找到1和n到两边距离的最大值即可。

代码：

/*--Codeforces 345 div2--Create by jiangyuzhu--2016/5/26*/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <stack>using namespace std;typedef long long ll;#define sa(n) scanf("%d", &(n))#define pl(x) cout << #x << " " << x << endl#define mdzz cout<<"mdzz"<<endl;int main (void){    int n;cin>>n;    int a;    int p1, p2;    for(int i = 1; i <= n; i++){        cin>>a;        if(a == 1) p1 = i;        if(a == n) p2 = i;    }    int res = max(p1 - 1, n - p1);    int res2 = max(p2 - 1, n - p2);    cout<<max(res, res2)<<endl;    return 0;}

B.Pyramid of Glasses

分析：

模拟一下倒水的过程。注意用double

代码：

/*--Codeforces 345 div2--Create by jiangyuzhu--2016/5/26*/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <stack>using namespace std;typedef long long ll;#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%I64d", &(n))#define pl(x) cout << #x << " " << x << endl#define mdzz cout<<"mdzz"<<endl;double a[10 + 5][10 + 5];int main (void){    int n, t;cin>>n>>t;    a[1][1] = t;    int cnt = 0;    for(int i = 1; i <= n; i++){        for(int j = 1; j <= i; j++){            if(a[i][j] >= 1){                a[i + 1][j] += (a[i][j] - 1) / 2.0;                a[i + 1][j + 1] += (a[i][j] - 1) / 2.0;                cnt++;            }        }    }    cout<<cnt<<endl;    return 0;}

C. Vasya and String

题意：

给定序列，可以任意更换k个字符，求更换后可以得到的最长的连续的相同的字符的长度。

分析：

典型的双指针

代码：

/*--Codeforces 345 div2--Create by jiangyuzhu--2016/5/26*/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <stack>using namespace std;typedef long long ll;#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%I64d", &(n))#define pl(x) cout << #x << " " << x << endl#define pr(x) cout << #x << " " << x << ' '#define mdzz cout<<"mdzz"<<endl;map<char, int>cnt;int main (void){    int n, k;cin>>n>>k;    string s;cin>>s;    int maxx = 0;    int len = s.length();    int l = 0, r = 1;    cnt[s[0]]++;    int res = 1;    while(l < r && r < len){        cnt[s[r]]++;        if(cnt[s[r]] > maxx){            maxx = cnt[s[r]];        }        if(r - l - maxx + 1<= k){            res = max(res, r - l + 1);        }        if(r - l - maxx + 1 > k){            cnt[s[l]]--;l++;        }        r++;    }    cout<<res<<endl;    return 0;}

D. Theseus and labyrinth

题意：

不同字符代表不同门的朝向，每秒钟在每块地方可以选择按钮让所有方块顺时针选择90度或者选择向相邻的方块中走。当且仅当两个门朝向相对时，才能从一个门进入另一个门，给定初始地点和终点，问最短时间。

分析：

一道很有意思的bfs。
我们用一个四位的二进制数来表示字符对应的门的情况，这样处理起来就简单了很多。结点保存坐标，时间和旋转的方向。那么对于某个固定的前进方向，我们只要判断旋转之后是否有门对着这个方向即可。

代码：

/*--Codeforces 345 div2--Create by jiangyuzhu--2016/5/26*/#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<queue>using namespace std;#define pr(x) cout << #x << ": " << x << "  "#define pl(x) cout << #x << ": " << x << endl;#define sa(x) scanf("%d",&(x))#define sal(x) scanf("%I64d",&(x))#define mdzz cout<<"mdzz"<<endl;const int maxn = 1e3 + 5, oo = 0x3f3f3f3f;char s[maxn][maxn];int a[maxn][maxn];int n, m;int x1, y1, x2, y2;struct NODE{int x; int y; int dist;int dir;};int vis[maxn][maxn][4];int dx[4] = {-1, 0, 1, 0};int dy[4] = {0, 1, 0, -1};int ans;int bfs(){    queue<NODE>q;    q.push((NODE){x1, y1, 0, 0});    vis[x1][y1][0] = 1;    while(!q.empty()){        NODE t = q.front();q.pop();        int xx = t.x, yy = t.y, dir = t.dir;        if(xx == x2 && yy == y2) return t.dist;        for(int i = 0; i < 4; i++){            int nx = xx + dx[i], ny = yy + dy[i];            if(nx >= n || nx < 0 || ny >= m|| ny < 0||vis[nx][ny][dir]) continue;            int aa = a[xx][yy] >> ((i + 4 - dir) % 4) & 1;            int bb = a[nx][ny] >> ((i + 6 - dir) % 4) & 1;            if(aa == 1 && bb == 1){                vis[nx][ny][dir] = 1;                q.push((NODE){nx, ny, t.dist + 1, dir});            }        }        int ndir = (dir + 1) % 4;        if(!vis[xx][yy][ndir]){            vis[xx][yy][ndir] = 1;            q.push((NODE){xx, yy, t.dist + 1, ndir});        }    }    return -1;}int main (void){    sa(n);sa(m);    for(int i = 0; i < n; i++){        scanf("%s",s[i]);        for(int j = 0; j < m; j++){            if(s[i][j] == '+') a[i][j] = 15;            if(s[i][j] == '-') a[i][j] = 10;            if(s[i][j] == '|') a[i][j] = 5;            if(s[i][j] == '^') a[i][j] = 1;            if(s[i][j] == '>') a[i][j] = 2;            if(s[i][j] == '<') a[i][j] = 8;            if(s[i][j] == 'v') a[i][j] = 4;            if(s[i][j] == 'L') a[i][j] = 7;            if(s[i][j] == 'R') a[i][j] = 13;            if(s[i][j] == 'U') a[i][j] = 14;            if(s[i][j] == 'D') a[i][j] = 11;            if(s[i][j] == '*') a[i][j] = 0;        }    }    sa(x1);sa(y1);sa(x2);sa(y2);    x1--;x2--;y1--;y2--;    printf("%d\n", bfs());    return 0;}

E. The Last Fight Between Human and AI

题意：

人和AI轮流向形如P(x) = anxn + an − 1xn − 1 + ... + a1x + a0 的多项式中填系数，AI和人都会选择最优决策。AI先手。
得到的多项式除以x−k可以得到另一个多项式的话，人赢。给定当前局面（可能是中间的局面），问最后谁赢。

分析：

首先明确：(x−k)|F(x)等价于F(k)==0
然后我们分情况讨论：

• k=0，只要看a0能否等于0即可。如果a0未填，并且正好轮到AI，那人肯定会输。
• k!=0，如果已经填满，只要判断F(k)是否等于0即可。否则需要判断最后是谁在填。当剩下最后一个系数ai未填时，无论之前填了什么数字，当我们取x=k的时候，原不等式即可转化为t∗k+ai∗ki，很明显，必定存在一个ai使得该不等式等于0。若是人，选择正确的ai，人赢。反之，AI任选其他系数，人输。所以只需看最后一个玩家是人还是AI即可。

代码：

/*--Codeforces 345 div2 E--Create by jiangyuzhu--2016/5/27*/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <stack>using namespace std;typedef long long ll;#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%I64d", &(n))#define pl(x) cout << #x << " " << x << endl#define pr(x) cout << #x << " " << x << ' 'const int maxn = 10 + 5, maxm = 1e5 + 5, oo = 0x3f3f3f3f, mod1 = 1e9 + 7, mod2 = 1e9 - 7;char s[maxn];int a[maxm];int cnt, k, n;bool solve(int fir){    if(k == 0){        if(a[0] == oo) return fir == 2;        else return a[0] == 0;    }    if(cnt == 0){        ll res = 0;        ll res2 = 0;        for(int i = n; i >= 0; i--){            res = (res * k % mod1 + a[i]) % mod1;            res2 = (res2 * k % mod2 + a[i]) % mod2;        }        return res == 0 && res2 == 0;    }else{        if((n + 1) & 1) return false;        else return true;    }}int main (void){    sa(n);sa(k);    cnt = 0;    for(int i = 0; i < n + 1; i++){        scanf("%s", s);        if(s[0] == '?'){            a[i] = oo;            cnt++;        }        else{            bool flag = true;            int len = strlen(s);            for(int j = 0; j < len; j++){                 if(s[j] == '-') {flag = false;continue;}                 a[i] = a[i] * 10 + s[j] - '0';            }           if(!flag) a[i] = -a[i];        }    }    int fir;    if((n + 1 - cnt) & 1)  fir = 2;//人    else fir = 1;//机器    if(solve(fir)) puts("Yes");    else puts("No");    return 0;}