2017 JUST Programming Contest 2.0

B. So You Think You Can Count?

设dp[i]表示以i为结尾的方案数，每个位置最多往前扫10位

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 const int mod = 1e9+;
 ll dp[maxn];//dp[i]表示以i结尾的方案数。
 int num[];
 char s[maxn];
 int main()
 {
     int n;
     cin >> n;
     cin >> s + ;
     dp[] = ;
     for(int i = ;i <= n;i++)
     {
         memset(num, ,sizeof(num));
         for(int j = i;j > ;j--)
         {
             num[s[j]-'']++;
             if(num[s[j]-''] > ) break;
             dp[i] = (dp[i] + dp[j-])%mod;
         }
     }
     cout << dp[n] << endl;
     return ;
 }

C. MRT Map

思路：存下每个字符串26个字母出现的次数，然后建边的时候计算一遍权值，然后dijkstra就好了。

 #include<bits/stdc++.h>
 using namespace std;
 typedef pair<int,int> P;
 const int maxn = 5e5 + ;
 const int INF = 0x3f3f3f3f;
 int n, m;
 int num[maxn][];
 int cal(int u, int v)
 {
     int ans = ;
     for(int i = ;i < ;i++){
         if(num[u][i] && num[v][i])
             ans++;
     }
     return ans;
 }
 struct edge{
     int to;
     int cost;
 }e;
 vector <edge> G[maxn];
 int d[maxn];
 void add(int u, int v)
 {
     e.to = v;
     e.cost = cal(u,v);
     G[u].push_back(e);
 }
 void dijkstra(int s)
 {
     priority_queue<P,vector<P>,greater<P> > que;
     fill(d ,d + n + ,INF);
     d[s] = ;
     que.push(P(,s));
     while(!que.empty())
     {
         P p= que.top();que.pop();
         int v = p.second;
         if(d[v] < p.first) continue;
         for(int i = ;i < G[v].size();i++)
         {
             edge e = G[v][i];;
             if(d[e.to] > d[v] +e.cost)
             {
                 d[e.to]= d[v] + e.cost;
                 que.push(P(d[e.to],e.to));
             }
         }
     }
 }
 int main()
 {
     std::ios::sync_with_stdio(false);
     cin >> n >> m;
     string a;
     for(int i = ;i <= n;i++)
     {
         cin >> a;
         int len = a.size();
         for(int j = ;j < len;j++){
             if(a[j] >= 'a' && a[j] <= 'z') num[i][a[j] - 'a']++;
             else num[i][a[j] - 'A']++;
         }

     }
     for(int i = ;i <= m;i++)
         {
             int u, v;
             cin >> u >> v;
             add(u, v);
             add(v, u);
         }
     int s, t;
     cin >> s >> t;
     dijkstra(s);
     cout << d[t] << endl;
     return ;
 }

D. Husam's Bug跑

签到题1

 #include<bits/stdc++.h>
 using namespace std;
 int main()
 {
     std::ios::sync_with_stdio(false);
     int t;
     cin >> t;
     while(t--)
     {
         string s;
         int cnt = , num = , sum = ;//字母 数字 特殊符号
         cin >> s;
         for(int i = ;i < s.size();i++)
         {
             if(s[i] >= 'a' && s[i] <= 'z') cnt++;
             if(s[i] >= 'A' && s[i] <= 'Z') cnt++;
             if(s[i] >= '' && s[i] <= '') num++;
             if(s[i] == '@' || s[i] == '?' || s[i] == '!') sum++;
         }
         if(cnt < ){
             cout << "The last character must be a letter." << endl;
         }
         else if(num < ){
             cout << "The last character must be a digit." << endl;
         }
         else if(sum < ){
             cout << "The last character must be a symbol." << endl;
         }
         else cout << "The last character can be any type." << endl;
     }
     return ;
 }

---

E. Abdalrahman Ali Bugs

思路：容易得出X是在 2 - 最大出现次数之间，直接暴力枚举所有答案取最优解。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll num[];
 ll check(ll x)
 {
     ll ans = ;
     for(int i = ;i < ;i++){
         ans += (num[i] % x) * num[i];
     }
     return ans;
 }
 int main()
 {
     std::ios::sync_with_stdio(false);
     string a;
     cin >> a;
     ll mx = ;
     for(int i = ;i < a.size();i++)
     {
         int t = a[i] - 'a';
         num[t]++;
         mx = max(num[t],mx);
     }
     ll x = ;
     ll ans = check(x);
     for(ll i = ;i <= mx;i++){
         if(check(i) < ans){
             x = i;
             ans = check(i);
         }
     }
     cout << x << endl;
     return ;
 }

---

F. Certifications

思路：在a数组中二分一个大于X的最小值，找不到则输出那一串。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 ll a[maxn];
 const ll INF = 1e18;
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n, m;
     cin >> n;
     for(int i = ;i < n;i++){
         cin >> a[i];
     }
     sort(a, a + n);
     a[n] = INF;
     cin >> m;
     while(m--)
     {
         int x;
         cin >> x;
         int l = , r = n, o = ;
         while(l <= r){
             int mid = (l + r) / ;
             if(a[mid] >= x) r = mid - , o = mid;
             else l = mid + ;
            // cout << o << endl;
         }
         if(a[o] == INF) cout << "Dr. Samer cannot take any offer :(." << endl;
         else cout << a[o] << endl;
     }
     return ;
 }

---

G. In the Chairman's office

qaq签到题2。

---

H. Give Me This Pizza

思路：右边最近的较大值，这不是很白的一个单调栈么，暑假队内赛第一场的A，相似度80%。

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn = 1e5 + ;
 int st[maxn], ans[maxn], a[maxn];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin >> n;
     memset(ans,-,sizeof(ans));
     for(int i = ;i < n;i++) cin >> a[i];
     int cnt = ;
     for(int i = ;i < n;i++)
     {
         while(cnt >  && a[st[cnt]] < a[i]){
             ans[st[cnt--]] = i;
         }
         st[++cnt] = i;
     }
     for(int i = ;i < n;i++){
         if(ans[i] == -) cout << ans[i] << " ";
         else cout << a[ans[i]] << " ";
     }
     return ;
 }

---

I. Husam and the Broken Present 1

思路：对角线元素是平方和，对其开根号即可。

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn = 1e5 + ;
 int ans[maxn];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin >> n;
     int a;
     for(int i = ;i < n;i++){
         for(int j = ;j < n;j++){
             cin >> a;
             if(i == j) ans[i] = sqrt(a);
         }
     }
     for(int i = ;i < n;i++){
         if(i == n - ) cout << ans[i] << endl;
         else cout << ans[i] << " ";
     }
     return ;
 }

---

K.Counting Time

思路：可以很暴力的一道题，9！枚举出3*3矩阵所有状态，判断和题目所给的矩阵非0位相不相等并且满不满足题意。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 int mp[][];
 int mp2[][];
 int a[];
 int vis[];
 int ans;
 int dx[] = {, , , -, , , -, -};
 int dy[] = {-, , , , -, , , -};

 bool check(int x, int y){
     if(x >=  && x <  && y >=  && y < ) return true;
     else return false;
 }
 bool check1(int x, int y,char val)
 {
     if(mp2[x][y] == ) return true;
     for(int i = ;i < ;i++)
     {
         int nx = x + dx[i];
         int ny = y + dy[i];
         if(check(nx, ny) && mp2[nx][ny] == val + ) return true;
     }
     return false;
 }
 bool check3(){
     for(int i = ;i < ;i++){
         for(int j = ;j < ;j++){
             if(!check1(i, j, mp2[i][j])) return false;
         }
     }
     return true;
 }
 int check2()
 {
     for(int i = ;i < ;i++){
         int x = i / ;
         int y = i % ;
         mp2[x][y] = a[i];
         if(mp[x][y] != a[i] && mp[x][y] != ) return ;
     }
     if(check3()) return ;
     else return ;
 }
 void dfs(int pos)
 {
     if(pos >= ){
         ans += check2();
         return;
     }
     for(int i = ;i <= ;i++)
     {
         if(!vis[i]) {
             a[pos] = i;
             vis[i] = ;
             dfs(pos + );
             vis[i] = ;
         }
     }
 }
 int main()
 {
     std::ios::sync_with_stdio(false);
     char c;
     for(int i = ;i < ;i++){
         for(int j = ;j < ;j++)
         {
             cin >> c;
             int t = c - '';
             mp[i][j] = t;
         }
     }
     dfs();
     cout << ans << endl;
     return ;
 }