2017 Russian Code Cup (RCC 17), Final Round

2017 Russian Code Cup (RCC 17), Final Round

A Set Theory

思路:原题转换一下就是找一个b数组,使得b数组任意两个数的差值都和a数组任意两个数的差值相等

根据题目数据范围, 肯定可以构造一个1, 1+d, 1+2d, 1+3d, ... , 1+(n-1)*d的序列

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = , M = 1e6 + ;
int a[N];
bool vis[M];
int main() {
    int n, T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for (int i = ; i <= n; i++) scanf("%d", &a[i]);
        sort(a+, a++n);
        bool f = true;
        for (int i = ; i <= n; i++) {
            if(a[i] == a[i-]) {
                f = false;
                break;
            }
        }
        if(!f) {
            puts("NO");
            continue;
        }
        f = true;
        mem(vis, false);
        for (int i = ; i <= n; i++) {
            for (int j = i+; j <= n; j++) {
                vis[a[j] - a[i]] = true;
            }
        }
        for (int i = ; i <= ; i++) {
            if(!vis[i]) {
                int cnt = ;
                for (int j = i+i; j <= ; j += i) {
                    if(!vis[j]) cnt++;
                    else break;
                }
                if(cnt >= n-) {
                    f = false;
                    puts("YES");
                    for (int j = ; j <= n; j ++) printf("%d ", +(j-)*i);
                    printf("\n");
                    break;
                }
            }
        }
        if(f) puts("NO");
    }
    return ;
}

B Similar Words

思路:对于每一个字符串xy, 如果它删去最前面字母得到的后缀y是某个串的前缀的话,我们把xy -> y

这样就会形成一个内向树森林,然后就是求任意两个相邻点最多选一个的点集的最大大小,这个可以用树形dp实现

然后就是要用字符串hash把字符串映射到数,这道题卡单hash,要用双hash

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e6 + , base1 = , base2 = ;
const int MOD = 1e9 + ;
string s[N];
bool vis[N];
vector<int> g[N];
vector<pair<ULL, ULL>> vc;
int dp[N][];
int dfs(int u, int o, int st) {
    if(~dp[u][st]) return dp[u][st];
    if(vis[u]) vis[u] = false;
    int ans = ;
    if(st) ans = ;
    for (int v : g[u]) {
        if(v != o) {
            if(st == ) ans += dfs(v, u, );
            else ans += max(dfs(v, u, ), dfs(v, u, ));
        }
    }
    return dp[u][st] = ans;
}
int main() {
    fio;
    int T, n;
    cin >> T;
    while(T--) {
        cin >> n;
        for (int i = ; i < n; i++) cin >> s[i];
        vc.clear();
        for (int i = ; i < n; i++) {
            ULL pre1 = , pre2 = ;
            for (int j = ; j < s[i].size(); j++) {
                pre1 = (pre1*base1 + s[i][j]-'a'+) % MOD;
                pre2 = pre2*base2 + s[i][j]-'a'+;
                vc.pb({pre1, pre2});
            }
        }
        sort(vc.begin(), vc.end());
        for (int i = ; i < n; i++) {
            ULL pre1 = , nxt1 = , pre2 = , nxt2 = ;
            for (int j = ; j < s[i].size(); j++) {
                pre1 = (pre1*base1 + s[i][j]-'a'+) % MOD;
                pre2 = pre2*base2 + s[i][j]-'a'+;
                if(j == ) {
                    int t = lower_bound(vc.begin(), vc.end(), pair<ULL,ULL>{pre1, pre2}) -vc.begin();
                    if(!vis[t]) {
                        vis[t] = true;
                    }
                }
                else {
                    nxt1 = (nxt1*base1 + s[i][j]-'a'+) % MOD;
                    nxt2 = nxt2*base2 + s[i][j]-'a'+;
                    int t = lower_bound(vc.begin(), vc.end(), pair<ULL,ULL>{pre1, pre2}) - vc.begin();
                    if(!vis[t]) {
                        vis[t] = true;
                        int tt = lower_bound(vc.begin(), vc.end(), pair<ULL,ULL>{nxt1, nxt2}) - vc.begin();
                        if(tt != vc.size() && vc[tt] == pair<ULL,ULL>{nxt1, nxt2}) {
                            g[t].pb(tt);
                            g[tt].pb(t);
                        }
                    }
                }
            }
        }
        int up = vc.size();
        for (int i = ; i < up; i++) dp[i][] = dp[i][] = -;
        int ans = ;
        for (int i = ; i < up; i++) {
            if(vis[i]) {
                ans += max(dfs(i, i, ), dfs(i, i, ));
            }
        }
        cout << ans << endl;
        for (int i = ; i < up; i++) g[i].clear(), vis[i] = false;
    }
    return ;
}

C Eleventh Birthday

D Masha and Cactus

E Satellites

F To Play or not to Play