#ctpc

This is usually a good combination. Both signs tend to have a strong emotional bond and a good understanding of each other. These two signs have a lot in common, and both long for security and warmth in relationships. Usually, they form a very tightly knit family. They both appreciate a solid home base and the material comforts of domestic life. Cancer, a water sign, is a born nurturer, and takes excellent care of the Taurus child. Both parent and child enjoy being at home, love each other`s company. But problems could arise if the Taurus child becomes stubborn causing the Cancer parent to sulk. As a result, this child needs to understand the parent`s emotional sensitivity, and the parent needs to be more open about his/her needs instead of relying on emotional blackmail. The Taurus-Cancer relationship is a solid, steady and reliable union. Both are dependable and strongly oriented toward family life. A mutual love of security and the home, makes theirs an ideal family relationship.

#include <iostream> #include <vector> using namespace std; const int MAX = 10000000; const int MOD = 1000000; bool p[MAX+1]; int main() { int n; cin >> n; vector<int> pr; for (int i=0; i<=MAX; i++) p[i]=1; p[0] = p[1] = 0; for (int i=2; i*i<=MAX; i++) { if (p[i]) { for (int k=i+i; k<=MAX; k+=i) p[k] = 0; } } for (int i=2; i<=MAX; i++) if (p[i]) pr.push_back(i); for (int i=0; i<n; i++) { long long num=1; int c; cin >> c; bool f=0; for (int k=0; !f&&k<pr.size(); k++) { num *= pr[k]; num %= MOD; if (num == c) { f=1; break; } } cout << (f ? "Yes" : "No") << endl; } }

#include <iostream> using namespace std; typedef long long ll; const int MOD = 1000000; int main() { ll n; cin >> n; ll N=1; while ((1<<(N-1)) < n) { N++; } ll a[N]; a[0] = 1; for (ll i=1; i<N; i++) { a[i] = a[i-1] * 7; a[i] %= MOD; } ll ans = 0; for (int i=0; i<N; i++) if (n&(1<<i)) ans = (ans + a[i]) % MOD; cout << ans << endl; }

#include <iostream> #include <map> #include <string> #include <cstring> using namespace std; const int MAX = 1<<20; int main() { bool tab[20][20]; memset(tab, 0, sizeof(tab)); int n; cin >> n; string s; map<string, int> dic; for (int i=0; i<n; i++) { cin >> s; dic[s] = i; } int S; cin >> S; for (int i=0; i<S; i++) { string s1, s2; cin >> s1 >> s2; tab[ dic[s1] ][ dic[s2] ] = 1; } bool f=0; for (int i=0; i<n-1; i++) for (int k=i+1; k<n; k++) if (!(tab[i][k]^tab[k][i])) { f=1; break; } if (f) { cout << "No" << endl; return 0; } int fw[n][n]; for (int i=0; i<n; i++) for (int k=0; k<n; k++) if (i==k) fw[i][i] = 0; else if (tab[i][k]) fw[i][k] = 1; else fw[i][k] = MAX; for (int w=0; w<n; w++) for (int i=0; i<n; i++) for (int k=0; k<n; k++) fw[i][k] = min(fw[i][k], fw[i][w]+fw[w][k]); f=0; for (int i=0; !f&&i<n; i++) { for (int k=0; !f&&k<n; k++) { if (i==k) continue; if (fw[i][k] == MAX) { f=1; break; } } } cout << (f ? "No" : "Yes") << endl; }

#include <iostream> #include <cstring> using namespace std; int dx[] = { 1, 0, -1, 0 }; int dy[] = { 0, 1, 0, -1 }; int main() { int H, W, K; cin >> W >> H >> K; int X[K], Y[K], L[K]; for (int i=0; i<K; i++) cin >> X[i] >> Y[i] >> L[i]; bool tab[H][W]; memset(tab, 0, sizeof(tab)); for (int i=0; i<K; i++) { bool f=1; for (int n=1; f; n++) { f=0; int siz = 2*n*(L[i]+1)-1; for (int x=X[i]-siz/2; x<=X[i]+siz/2; x++) { int y1=Y[i]-siz/2, y2=Y[i]+siz/2; if (0<=x&&x<W&&0<=y1&&y1<H) { tab[y1][x] ^= true; f=1; } if (0<=x&&x<W&&0<=y2&&y2<H) { tab[y2][x] ^= true; f=1; } } for (int y=Y[i]-siz/2+1; y<Y[i]+siz/2; y++) { int x1=X[i]-siz/2, x2=X[i]+siz/2; if (0<=x1&&x1<W&&0<=y&&y<H) { tab[y][x1] ^= true; f=1; } if (0<=x2&&x2<W&&0<=y&&y<H) { tab[y][x2] ^= true; f=1; } } } } for (int y=0; y<H; y++) { for (int x=0; x<W; x++) cout << (tab[y][x] ? '#' : '.'); cout << endl; } }

#include <iostream> using namespace std; int main() { int s = 0; int m = 0; int t; for (int i=0; i<4; i++) { cin >> t; s += t; m = max(m, t); } s += m; s /= 5; cout << (s>=60 ? "Yes" : "No") << endl; }