[Codeforces]Good Bye 2017

A - New Year and Counting Cards

#pragma comment(linker, "/STACK:102400000,102400000")

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<functional>

#include<math.h>

//#include<bits/stdc++.h>

using namespace std;

typedef long long lint;

typedef vector<int> VI;

typedef pair<int, int> PII;

typedef queue<int> QI;

void makedata() {

    freopen("input.txt", "w", stdout);

    fclose(stdout);

}

int a[];

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    //makedata();

    std::ios::sync_with_stdio(), cin.tie();

    string s;

    cin >> s;

    int n = s.size();

    for (int i = ; i < n; i++) a[s[i]]++;

    cout << (a[''] + a[''] + a[''] + a[''] + a[''] + a['a'] + a['e'] + a['i'] + a['o'] + a['u']) << endl;

    return ;

}

B - New Year and Buggy Bot

#pragma comment(linker, "/STACK:102400000,102400000")

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<functional>

#include<math.h>

//#include<bits/stdc++.h>

using namespace std;

typedef long long lint;

typedef vector<int> VI;

typedef pair<int, int> PII;

typedef queue<int> QI;

void makedata() {

    freopen("input.txt", "w", stdout);

    fclose(stdout);

}

string mp[], s;

int d[], op[];

const int dx[] = {-, , , };

const int dy[] = {, , , -};

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    //makedata();

    std::ios::sync_with_stdio(), cin.tie();

    int n, m;

    cin >> n >> m;

    for (int i = ; i < n; i++) cin >> mp[i];

    cin >> s;

    int l = s.size();

    for (int i = ; i < l; i++) op[i] = s[i] - '';

    int sx, sy;

    for (int i = ; i < n; i++) {

        for (int j = ; j < m; j++) {

            if (mp[i][j] == 'S') {

                sx = i;

                sy = j;

            }

        }

    }

    int ans = ;

    bool f[];

    for (d[] = ; d[] < ; d[]++) {

        for (d[] = ; d[] < ; d[]++) {

            for (d[] = ; d[] < ; d[]++) {

                for (d[] = ; d[] < ; d[]++) {

                    memset(f, false, sizeof(f));

                    f[d[]] = true, f[d[]] = true, f[d[]] = true, f[d[]] = true;

                    if (f[] && f[] && f[] && f[]) {

                        int x = sx, y = sy, j;

                        bool ok = false;

                        for (int i = ; i < l; i++) {

                            j = d[op[i]];

                            x += dx[j], y += dy[j];

                            if (x <  || x >= n || y <  || y >= m) break;

                            if (mp[x][y] == '#') break;

                            if (mp[x][y] == 'E') {

                                ok = true;

                                break;

                            }

                        }

                        if (ok) ans++;

                    }

                }

            }

        }

    }

    cout << ans << endl;

    return ;

}

C - New Year and Curling

#pragma comment(linker, "/STACK:102400000,102400000")

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<functional>

#include<math.h>

//#include<bits/stdc++.h>

using namespace std;

typedef long long lint;

typedef vector<int> VI;

typedef pair<int, int> PII;

typedef queue<int> QI;

void makedata() {

    freopen("input.txt", "w", stdout);

    fclose(stdout);

}

class SAM {

private:

    int Last, sz;

    vector<int []> ch;

    vector<int> fa, maxlen;

public:

    void init() {

        sz = ;

        ch.clear();

    }

    void add(int x) {

        int np = ++sz, p = Last;

        Last = np;

        memset(ch[np], , sizeof(ch[np]));

        maxlen[np] = maxlen[p] + ;

        while (p && !ch[p][x]) ch[p][x] = np, p = fa[p];

        if (!p) fa[np] = ;

        else {

            int q = ch[p][x];

            if (maxlen[p] +  == maxlen[q]) fa[np] = q;

            else {

                int nq = ++sz;

                memcpy(ch[nq], ch[q], sizeof(ch[q]));

                maxlen[nq] = maxlen[p] + ;

                fa[nq] = fa[q];

                fa[q] = fa[np] = nq;

                while (p && ch[p][x] == q) ch[p][x] = nq, p = fa[p];

            }

        }

    }

};

double x[], y[];

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    //makedata();

    //std::ios::sync_with_stdio(0), cin.tie(0);

    int n;

    double r, h;

    //cin >> n >> r;

    scanf("%d%lf", &n, &r);

    for (int i = ; i < n; i++) {

        //cin >> x[i];

        scanf("%lf", &x[i]);

        y[i] = r;

        for (int j = ; j < i; j++) {

            if (fabs(x[i] - x[j]) <=  * r) {

                h = sqrt( * r * r - (x[i] - x[j]) * (x[i] - x[j])) + y[j];

                if (y[i] < h) y[i] = h;

            }

        }

    }

    for (int i = ; i < n; i++) printf("%.8lf ", y[i]); //cout << y[i] << ' ';

    return ;

}

D - New Year and Arbitrary Arrangement

#include <bits/stdc++.h>

using ll = long long;

using ld = long double;

using namespace std;

const int MOD =  *  *  + ;

ll pw(ll a, ll b = MOD - ) {

    if (b == ) {

        return ;

    }

    ll v = pw(a, b / );

    v = (v * v) % MOD;

    if (b & ) {

        v = (v * a) % MOD;

    }

    return v;

}

constexpr int MAXN = ;

ll f[MAXN][MAXN];

int main() {

#ifdef PAUNSVOKNO

    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

#endif

    ios_base::sync_with_stdio(false); cout.setf(ios::fixed); cout.precision(); cout.tie(nullptr); cin.tie(nullptr);

    ll k, pa, pb;

    cin >> k >> pa >> pb;

    ll s = pw(pa + pb);

    pa = (pa * s) % MOD;

    pb = (pb * s) % MOD;

    ll ans = ;

    f[][] = ;

    ll last = ( - pa + MOD) % MOD;

    last = pw(last, MOD - );

    last = (last -  + MOD) % MOD;

    for (int i = ; i <= k; ++i) {

        for (int j = ; j <= k; ++j) {

            if (i + j >= k) {

                ll val = j + i + last;

                ans = (ans + f[i][j] * val) % MOD;

            } else {

                f[i + ][j] = (f[i + ][j] + f[i][j] * pa) % MOD;

                f[i][j + i] = (f[i][j + i] + f[i][j] * pb) % MOD;

            }

        }

    }

    cout << ans << "\n";

}

E - New Year and Entity Enumeration

#include<bits/stdc++.h>

#define ll long long

#define mod 1000000007

using namespace std;

ll m,n,c[][],f[],b[],x,ans=;

map<ll,ll>mp;

int main(){

    for (cin>>m>>n;n--;)

        for (int i=;i<m;++i)

            scanf("%1lld",&x), b[i]|=x<<n;

    for (int i=;i<m;++i) ++mp[b[i]];

    for (int i=;i<=m;++i)

        for (int j=;j<=i;++j)

            c[i][j]= j? (c[i-][j-]+c[i-][j])%mod: ;

    f[]=f[]=;

    for (int i=;i<=m;++i)

        for (int j=;j<i;++j)

            f[i]=(f[i]+c[i-][j]*f[j])%mod;

    for (auto o:mp) ans=(ans*f[o.second])%mod;

    cout<<ans;

}

F - New Year and Rainbow Roads

#pragma comment(linker, "/STACK:102400000,102400000")

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<functional>

#include<math.h>

//#include<bits/stdc++.h>

using namespace std;

typedef long long lint;

typedef vector<int> VI;

typedef pair<int, int> PII;

typedef queue<int> QI;

void makedata() {

    freopen("input.txt", "w", stdout);

    fclose(stdout);

}

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    //makedata();

    std::ios::sync_with_stdio(), cin.tie();

    int n;

    char ch;

    lint pb = , pr = , pg = , mr = , mb = , ans = , x;

    cin >> n;

    for (int i = ; i < n; i++) {

        cin >> x >> ch;

        if (ch == 'R' || ch == 'G') {

            if (pr) {

                ans += x - pr;

                mr = max(mr, x - pr);

            }

            pr = x;

        }

        if (ch == 'B' || ch == 'G') {

            if (pb) {

                ans += x - pb;

                mb = max(mb, x - pb);

            }

            pb = x;

        }

        if (ch == 'G') {

            if (pg) ans += min(0LL, x - pg - mr - mb);

            pg = x;

            mr = mb = ;

        }

    }

    cout << ans << endl;

    return ;

}

G - New Year and Original Order

#pragma comment(linker, "/STACK:102400000,102400000")

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

#include<algorithm>

#include<iostream>

#include<map>

#include<queue>

#include<stack>

#include<string>

#include<functional>

#include<math.h>

//#include<bits/stdc++.h>

using namespace std;

typedef long long lint;

typedef vector<int> VI;

typedef pair<int, int> PII;

typedef queue<int> QI;

void makedata() {

    freopen("input.txt", "w", stdout);

    fclose(stdout);

}

const int mod = ;

lint dp[][][][];

string s;

int main() {

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

#endif

    //makedata();

    std::ios::sync_with_stdio(), cin.tie();

    cin >> s;

    int n = s.size();

    reverse(s.begin(), s.end());

    memset(dp, , sizeof(dp));

    for (int i = ; i < ; i++) dp[][][i][] = ;

    for (int i = ; i < n; i++) {

        for (int j = ; j <= i; j++) {

            for (int k = ; k < ; k++) {

                for (int l = ; l < ; l++) {

                    for (int p = ; p < ; p++) {

                        (dp[i + ][j + (p >= k)][k][(s[i] - '' > p) || (l && (s[i] - '' == p))] += dp[i][j][k][l]) %= mod;

                    }

                }

            }

        }

    }

    lint ans = ;

    for (int i = ; i < ; i++) {

        lint t = ;

        for (int j = ; j <= n; j++, t = (t *  + ) % mod) ans += dp[n][j][i][] * t, ans %= mod;

    }

    cout << ans << endl;

    return ;

}

H - New Year and Boolean Bridges

#include<bits/stdc++.h>

#define N 50

#define fr(i,n) for(int i=1;i<=n;i++)

using namespace std;

int n,i,j,x,S,an,nv,id,to[N],c[N],b[N],F[N],sz[N];bool v[N][N];

char s[N][N];

int gf(int x){return F[x]==x?x:F[x]=gf(F[x]);}

void uni(int x,int y){

    x=gf(x);y=gf(y);

    if(x!=y)F[x]=y,sz[y]+=sz[x];

}

int main(){

    scanf("%d",&n);

    fr(i,n)F[i]=i,sz[i]=;

    fr(i,n){

        scanf("%s",s[i]+);

        fr(j,n)if(i!=j)

            if(s[i][j]=='A')uni(i,j);

    }

    fr(i,n)if(gf(i)==i&&sz[i]>)to[i]=++id;

    fr(i,n)fr(j,n)if(gf(i)==gf(j)&&s[i][j]=='X')return puts("-1"),;

    fr(i,n)fr(j,n)if(s[i][j]=='X')v[to[gf(i)]][to[gf(j)]]=;

    fr(i,id)b[i]=i;an=n;

    fr(T,){

        random_shuffle(b+,b+id+);nv=;

        fr(i,id){

            x=b[i];S=;

            fr(j,id)if(v[x][j])S|=<<c[j];

            for(j=;;j++)if(!(S>>j&)){

                c[x]=j;

                break;

            }

            nv=max(nv,c[x]);

        }

        fr(i,id)c[i]=;

        an=min(an,nv);

    }

    printf("%d\n",n-+an);

}

[Codeforces]Good Bye 2017的更多相关文章

Codeforces Good Bye 2017 908F F. New Year and Rainbow Roads
题 OvO http://codeforces.com/contest/908/problem/F CF 908F 解需要注意细节的模拟题. 如果三种颜色都存在,则记每两个相邻的G组成一个段,对每个 ...
codeforces Good bye 2016 E 线段树维护dp区间合并
codeforces Good bye 2016 E 线段树维护dp区间合并题目大意:给你一个字符串,范围为‘0’~'9',定义一个ugly的串,即串中的子串不能有2016,但是一定要有2017,问 ...
Good Bye 2017 A B C
Good Bye 2017 A New Year and Counting Cards 题目链接: http://codeforces.com/contest/908/problem/A 思路: 如果 ...
Good Bye 2017(送命场)
9815人数场,9500+围观神仙打架...断断续续打Codeforces也快有一年啦,第一次打Good Bye场,满怀前排膜tourist的心愿参加了这场送命场,虽然没看到tourist.不过还是得 ...
Codeforces:Good Bye 2018（题解）
Good Bye 2018! 题目链接:https://codeforces.com/contest/1091 A. New Year and the Christmas Ornament 题意: 给 ...
codeforces Good Bye 2015 B. New Year and Old Property
题目链接:http://codeforces.com/problemset/problem/611/B 题目意思:就是在 [a, b] 这个范围内(1 ≤ a ≤ b ≤ 10^18)统计出符合二进制 ...
Codeforces Good Bye 2015 D. New Year and Ancient Prophecy 后缀数组树状数组 dp
D. New Year and Ancient Prophecy 题目连接: http://www.codeforces.com/contest/611/problem/C Description L ...
Codeforces Good Bye 2015 C. New Year and Domino 前缀和
C. New Year and Domino 题目连接: http://www.codeforces.com/contest/611/problem/C Description They say &q ...
Codeforces Good bye 2015 B. New Year and Old Property dfs 数位DP
B. New Year and Old Property 题目连接: http://www.codeforces.com/contest/611/problem/B Description The y ...

随机推荐

javaee字符文件的复制
package Zy; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.FileWrit ...
UI Testing
UI Test能帮助我们去验证一些UI元素的属性和状态.Apple 在 Xcode 7 中新加入了一套 UI Testing 的工具,其目的就是解决自动化UI测试这个问题.新的 UI Testing ...
python 疑难杂症
1.getpass模块 :Pycharm不支持getpass模块,使用terminal可运行,但是getpass中文提示显示乱码?
单行函数、表连接(day02)
回顾: 1.数据库介绍 sql: dql: select dml: insert delete update ddl: create drop alter tcl: commit rollback s ...
HDU2866 Special Prime
题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=2866 题意:在区间[2,L]内,有多少个素数p,满足方程有解. 分析: 原方程变为: n^(b-1) ...
java实现根据高德地图API接口进行地址位置解析，将地址转化为经纬度
原创文章,转载请注明,欢迎评论和更改. 1,所需额外ar包,import net.sf.json.JSONObject; 2,完整源代码代码 package com.travel.util; impo ...
JS移动客户端--触屏滑动事件及js手机拖拽效果
移动端触屏滑动的效果其实就是图片轮播,在PC的页面上很好实现,绑定click和mouseover等事件来完成.但是在移动设备上,要实现这种轮播的效果,就需要用到核心的touch事件.处理touch事件 ...
PHP统计当前在线用户数实例
HTML 我们在页面上放置一个显示当前在线人数的div#total以及一个用于展示访客地区分布的列表#onlinelist,默认我们在列表中放置一张与加载动画图片,后面我们用jQuery控制当鼠标滑向 ...
5、SpringBoot+MyBaits+Maven+Idea+pagehelper分页插件
1.为了我们平时方便开发,我们可以在同一个idea窗口创建多个项目模块,创建方式如下 2.项目中pom.xm文件的内容如下 <?xml version="1.0" encod ...
邓_ Phpcms·二次开发
PHPCMS V9产品介绍 PHPCMS V9(简称V9)采用PHP5+MYSQL做为技术基础进行开发.V9采用OOP(面向对象)方式进行基础运行框架搭建.模块化开发方式做为功能开发形式.框架易于功能 ...

[Codeforces]Good Bye 2017

[Codeforces]Good Bye 2017的更多相关文章

随机推荐

热门专题