Links: https://codeforces.com/contest/1228
A: attendance problems, violence found in the past from L to R.
1 #include<bits/stdc++.h> 2 using namespace std; 3 bool check(int x) 4 { 5 int vis[10] = {0}; 6 while(x){ 7 if(vis[x % 10]) return false; 8 vis[x % 10]++; 9 x /= 10; 10 } 11 return true; 12 } 13 int main() 14 { 15 std::ios::sync_with_stdio(false); 16 int l, r; 17 cin >> l >> r; 18 for(int i = l;i <= r;i++){ 19 if(check(i)){ 20 cout << i << endl; 21 return 0; 22 } 23 } 24 cout << -1 << endl; 25 return 0; 26 }
B: color position may be determined per row r [i] +1 and c [i] + 1, and outputs 0 if the conflict, otherwise the output block number 2 ^ uncertain color.
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e3 + 50; 4 const int mod = 1000000007; 5 typedef long long ll; 6 int r, c; 7 int mp[maxn][maxn]; 8 ll pow_mod(ll x, ll n){ 9 ll res = 1; 10 while(n){ 11 if(n & 1) res = res * x % mod; 12 x = x * x % mod; 13 n >>= 1; 14 } 15 return res; 16 } 17 int main() 18 { 19 std::ios::sync_with_stdio(false); 20 int n, m; 21 cin >> n >> m; 22 int ans = 0; 23 bool flag = false; 24 for(int i = 0;i < n;i++){ 25 cin >> r; 26 for(int j = 0;j < r;j++) 27 mp[i][j] = 1; 28 mp[i][r] = 2; 29 } 30 for(int i = 0;i < m;i++) 31 { 32 cin >> c; 33 for(int j = 0;j < c;j++){ 34 if(mp[j][i] && mp[j][i] != 1)flag = true; 35 else mp[j][i] = 1; 36 } 37 if(mp[c][i] && mp[c][i] != 2) flag = true; 38 else mp[c][i] = 2; 39 } 40 if(flag){ 41 cout << 0 << endl; 42 return 0; 43 } 44 for(int i = 0;i < n;i++) 45 for(int j = 0;j < m;j++) 46 if(!mp[i][j]) ans++; 47 cout << pow_mod(2, ans) % mod << endl; 48 return 0; 49 }
C: to find the product of n-th power of the prime factor in the occurrence of x!.
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int mod = 1e9 + 7; 5 vector<ll> vec; 6 void init(ll n) 7 { 8 for(ll i = 2;i * i <= n;i++){ 9 if(n % i == 0) 10 { 11 vec.push_back(i); 12 while(n % i == 0) n /= i; 13 } 14 } 15 if(n > 1) vec.push_back(n); 16 } 17 ll pow_mod(ll x, ll n){ 18 ll res = 1; 19 while(n){ 20 if(n & 1) res = res * x % mod; 21 x = x * x % mod; 22 n >>= 1; 23 } 24 return res % mod; 25 } 26 ll fact_pow(ll n, ll k) { 27 ll res = 0; 28 while (n) { 29 n /= k; 30 res += n; 31 } 32 return res; 33 } 34 int main() 35 { 36 std::ios::sync_with_stdio(false); 37 ll x, n; 38 cin >> x >> n; 39 init(x); 40 ll ans = 1; 41 for(int i = 0;i < vec.size();i++){ 42 ll k = fact_pow(n, vec[i]); 43 ans = ans * pow_mod(vec[i], k) % mod; 44 } 45 cout << ans % mod <<endl; 46 return 0; 47 }
D: third graph coloring, firstly a dye, and then all side connection 1 2 is not 3, the side edge connections is 1, 2 empathy and dyeing. Finally, whether the subject does not satisfy the conditions to the full.
. 1 #include <bits / STDC ++ H.> 2 the using namespace STD; . 3 const int MAXN = 4E5 + . 5 ; . 4 Vector < int > G [MAXN]; . 5 int du [MAXN]; . 6 int Color [MAXN]; . 7 int A [MAXN], B [MAXN]; . 8 int FAR [MAXN]; . 9 Vector < int > C [ . 4 ]; // save each division point in FIG. 10 int n-, m; . 11 void DFS ( int V) 12 is { 13 for(int i = 0;i < G[v].size();i++) 14 { 15 int u = G[v][i]; 16 if(color[u]) continue; 17 color[u] = 3; 18 C[3].push_back(u); 19 } 20 } 21 int main() 22 { 23 std::ios::sync_with_stdio(false); 24 cin >> n >> m; 25 //建图 26 for(int i = 1;i <= m;i++){ 27 cin >> a[i] >> b[i]; 28 G[a[i]].push_back(b[i]); 29 G[b[i]].push_back(a[i]); 30 du[a[i]]++; 31 du[b[i]]++; 32 } 33 //染色1. 34 color[1] = 1; 35 C[1].push_back(1); 36 far[1]= 1; 37 for(int i = 0;i < G[1].size();i++) far[ G[1][i] ] = 1; 38 for(int i = 1;i <= n;i++) 39 { 40 if(far[i] == 0) 41 { 42 color[i] = 1; 43 C[1].push_back(i); 44 } 45 } 46 //染色2 47 for(int i = 1;i <= n;i++) 48 { 49 if(!color[i]) 50 { 51 color[i] = 2; 52 C[2].push_back(i); 53 54 dfs(i);//染色3 55 } 56 } 57 if(C[1].size() + C[2].size() + C[3].size() != n) 58 { 59 cout << -1 << endl; 60 return 0 ; 61 is } 62 is IF (C [ . 1 ] .size () == 0 || C [ 2 ] .size () == 0 || C [ . 3 ] .size () == 0 ) 63 is { 64 COUT < <- . 1 << endl; 65 return 0 ; 66 } 67 for ( int I = . 1 ; I <= n-; I ++ ) 68 { 69 IF ! (du [I] = n-- C [Color [I]] size. ()) // check each point are not its point we have a set of edges 70 { 71 cout << -1 << endl; 72 return 0; 73 } 74 } 75 for(int i = 0;i < C[1].size();i++) far[C[1][i] ] = 1; 76 for(int i = 0;i < C[2].size();i++) far[C[2][i] ] = 2; 77 for(int i = 0;i < C[3].size();i++) far[C[3][i] ] = 3; 78 for(int i = 1;i <= m;i++){ 79 if(far[a[i]] == far[b[i]]) 80 { 81 cout << -1 << endl; 82 return 0; 83 } 84 } 85 for(int i = 1;i <= n;i++){ 86 cout << color[i] << " "; 87 } 88 cout << endl; 89 return 0; 90 }