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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5 + 50; 5 const int mod = 1e9+7; 6 ll dp[maxn];//dp[i]表示以i结尾的方案数。 7 int num[20]; 8 char s[maxn]; 9 int main() 10 { 11 int n; 12 cin >> n; 13 cin >> s + 1; 14 dp[0] = 1; 15 for(int i = 1;i <= n;i++) 16 { 17 memset(num, 0,sizeof(num)); 18 for(int j = i;j > 0;j--) 19 { 20 num[s[j]-'0']++; 21 if(num[s[j]-'0'] > 1) break; 22 dp[i] = (dp[i] + dp[j-1])%mod; 23 } 24 } 25 cout << dp[n] << endl; 26 return 0; 27 }
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef pair<int,int> P; 4 const int maxn = 5e5 + 50; 5 const int INF = 0x3f3f3f3f; 6 int n, m; 7 int num[maxn][26]; 8 int cal(int u, int v) 9 { 10 int ans = 0; 11 for(int i = 0;i < 26;i++){ 12 if(num[u][i] && num[v][i]) 13 ans++; 14 } 15 return ans; 16 } 17 struct edge{ 18 int to; 19 int cost; 20 }e; 21 vector <edge> G[maxn]; 22 int d[maxn]; 23 void add(int u, int v) 24 { 25 e.to = v; 26 e.cost = cal(u,v); 27 G[u].push_back(e); 28 } 29 void dijkstra(int s) 30 { 31 priority_queue<P,vector<P>,greater<P> > que; 32 fill(d ,d + n + 1,INF); 33 d[s] = 0; 34 que.push(P(0,s)); 35 while(!que.empty()) 36 { 37 P p= que.top();que.pop(); 38 int v = p.second; 39 if(d[v] < p.first) continue; 40 for(int i = 0;i < G[v].size();i++) 41 { 42 edge e = G[v][i];; 43 if(d[e.to] > d[v] +e.cost) 44 { 45 d[e.to]= d[v] + e.cost; 46 que.push(P(d[e.to],e.to)); 47 } 48 } 49 } 50 } 51 int main() 52 { 53 std::ios::sync_with_stdio(false); 54 cin >> n >> m; 55 string a; 56 for(int i = 1;i <= n;i++) 57 { 58 cin >> a; 59 int len = a.size(); 60 for(int j = 0;j < len;j++){ 61 if(a[j] >= 'a' && a[j] <= 'z') num[i][a[j] - 'a']++; 62 else num[i][a[j] - 'A']++; 63 } 64 65 } 66 for(int i = 1;i <= m;i++) 67 { 68 int u, v; 69 cin >> u >> v; 70 add(u, v); 71 add(v, u); 72 } 73 int s, t; 74 cin >> s >> t; 75 dijkstra(s); 76 cout << d[t] << endl; 77 return 0; 78 }
D. Husam's Bug跑
签到题1
1 #include<bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 std::ios::sync_with_stdio(false); 6 int t; 7 cin >> t; 8 while(t--) 9 { 10 string s; 11 int cnt = 0, num = 0, sum = 0;//字母 数字 特殊符号 12 cin >> s; 13 for(int i = 0;i < s.size();i++) 14 { 15 if(s[i] >= 'a' && s[i] <= 'z') cnt++; 16 if(s[i] >= 'A' && s[i] <= 'Z') cnt++; 17 if(s[i] >= '0' && s[i] <= '9') num++; 18 if(s[i] == '@' || s[i] == '?' || s[i] == '!') sum++; 19 } 20 if(cnt < 4){ 21 cout << "The last character must be a letter." << endl; 22 } 23 else if(num < 4){ 24 cout << "The last character must be a digit." << endl; 25 } 26 else if(sum < 2){ 27 cout << "The last character must be a symbol." << endl; 28 } 29 else cout << "The last character can be any type." << endl; 30 } 31 return 0; 32 }
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 ll num[27]; 5 ll check(ll x) 6 { 7 ll ans = 0; 8 for(int i = 0;i < 26;i++){ 9 ans += (num[i] % x) * num[i]; 10 } 11 return ans; 12 } 13 int main() 14 { 15 std::ios::sync_with_stdio(false); 16 string a; 17 cin >> a; 18 ll mx = 0; 19 for(int i = 0;i < a.size();i++) 20 { 21 int t = a[i] - 'a'; 22 num[t]++; 23 mx = max(num[t],mx); 24 } 25 ll x = 2; 26 ll ans = check(x); 27 for(ll i = 2;i <= mx;i++){ 28 if(check(i) < ans){ 29 x = i; 30 ans = check(i); 31 } 32 } 33 cout << x << endl; 34 return 0; 35 }
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5 + 50; 5 ll a[maxn]; 6 const ll INF = 1e18; 7 int main() 8 { 9 std::ios::sync_with_stdio(false); 10 int n, m; 11 cin >> n; 12 for(int i = 0;i < n;i++){ 13 cin >> a[i]; 14 } 15 sort(a, a + n); 16 a[n] = INF; 17 cin >> m; 18 while(m--) 19 { 20 int x; 21 cin >> x; 22 int l = 0, r = n, o = 0; 23 while(l <= r){ 24 int mid = (l + r) / 2; 25 if(a[mid] >= x) r = mid - 1, o = mid; 26 else l = mid + 1; 27 // cout << o << endl; 28 } 29 if(a[o] == INF) cout << "Dr. Samer cannot take any offer :(." << endl; 30 else cout << a[o] << endl; 31 } 32 return 0; 33 }
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%。
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 50; 4 int st[maxn], ans[maxn], a[maxn]; 5 int main() 6 { 7 std::ios::sync_with_stdio(false); 8 int n; 9 cin >> n; 10 memset(ans,-1,sizeof(ans)); 11 for(int i = 0;i < n;i++) cin >> a[i]; 12 int cnt = 0; 13 for(int i = 0;i < n;i++) 14 { 15 while(cnt > 0 && a[st[cnt]] < a[i]){ 16 ans[st[cnt--]] = i; 17 } 18 st[++cnt] = i; 19 } 20 for(int i = 0;i < n;i++){ 21 if(ans[i] == -1) cout << ans[i] << " "; 22 else cout << a[ans[i]] << " "; 23 } 24 return 0; 25 }
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 50; 4 int ans[maxn]; 5 int main() 6 { 7 std::ios::sync_with_stdio(false); 8 int n; 9 cin >> n; 10 int a; 11 for(int i = 0;i < n;i++){ 12 for(int j = 0;j < n;j++){ 13 cin >> a; 14 if(i == j) ans[i] = sqrt(a); 15 } 16 } 17 for(int i = 0;i < n;i++){ 18 if(i == n - 1) cout << ans[i] << endl; 19 else cout << ans[i] << " "; 20 } 21 return 0; 22 }
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。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int mp[10][10]; 5 int mp2[10][10]; 6 int a[10]; 7 int vis[10]; 8 int ans; 9 int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1}; 10 int dy[8] = {-1, 0, 1, 0, -1, 1, 1, -1}; 11 12 bool check(int x, int y){ 13 if(x >= 0 && x < 3 && y >= 0 && y < 3) return true; 14 else return false; 15 } 16 bool check1(int x, int y,char val) 17 { 18 if(mp2[x][y] == 9) return true; 19 for(int i = 0;i < 8;i++) 20 { 21 int nx = x + dx[i]; 22 int ny = y + dy[i]; 23 if(check(nx, ny) && mp2[nx][ny] == val + 1) return true; 24 } 25 return false; 26 } 27 bool check3(){ 28 for(int i = 0;i < 3;i++){ 29 for(int j = 0;j < 3;j++){ 30 if(!check1(i, j, mp2[i][j])) return false; 31 } 32 } 33 return true; 34 } 35 int check2() 36 { 37 for(int i = 0;i < 9;i++){ 38 int x = i / 3; 39 int y = i % 3; 40 mp2[x][y] = a[i]; 41 if(mp[x][y] != a[i] && mp[x][y] != 0) return 0; 42 } 43 if(check3()) return 1; 44 else return 0; 45 } 46 void dfs(int pos) 47 { 48 if(pos >= 9){ 49 ans += check2(); 50 return; 51 } 52 for(int i = 1;i <= 9;i++) 53 { 54 if(!vis[i]) { 55 a[pos] = i; 56 vis[i] = 1; 57 dfs(pos + 1); 58 vis[i] = 0; 59 } 60 } 61 } 62 int main() 63 { 64 std::ios::sync_with_stdio(false); 65 char c; 66 for(int i = 0;i < 3;i++){ 67 for(int j = 0;j < 3;j++) 68 { 69 cin >> c; 70 int t = c - '0'; 71 mp[i][j] = t; 72 } 73 } 74 dfs(0); 75 cout << ans << endl; 76 return 0; 77 }