牛客小白月赛4
经典暴力：由于不能构成三角形边长至少乘二，只用考虑最大的log个暴力
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 typedef long long LL; 5 LL a[maxn], id[maxn], vis[maxn], ans[maxn]; 6 7 bool cmp(LL i, LL j) { 8 return a[i] > a[j]; 9 } 10 11 int main() { 12 int n, q; 13 scanf("%d %d", &n, &q); 14 for(int i = 1; i <= n; ++i) scanf("%lld", a + i), id[i] = i; 15 sort(id + 1, id + 1 + n, cmp); 16 LL M = -1; 17 for(int i = 1; i <= min(35, n); ++i) { 18 vis[id[i]] = 1; 19 for(int j = i + 1; j <= min(35, n); ++j) { 20 for(int k = j + 1; k <= min(35, n); ++k) { 21 LL I = id[i], J = id[j], K = id[k]; 22 if(a[I] + a[J] > a[K] && a[J] + a[K] > a[I] && a[I] + a[K] > a[J]) 23 M = max(M, a[I] + a[J] + a[K]); 24 } 25 } 26 } 27 for(int o = 1; o <= n; ++o) { 28 if(!vis[o]) ans[o] = M; 29 else { 30 LL MM = -1; 31 for(int i = 1; i <= min(35, n); ++i) { 32 if(id[i] == o) continue; 33 for(int j = i + 1; j <= min(35, n); ++j) { 34 if(id[j] == o) continue; 35 for(int k = j + 1; k <= min(35, n); ++k) { 36 if(id[k] == o) continue; 37 LL I = id[i], J = id[j], K = id[k]; 38 if(a[I] + a[J] > a[K] && a[J] + a[K] > a[I] && a[I] + a[K] > a[J]) 39 MM = max(MM, a[I] + a[J] + a[K]); 40 } 41 } 42 } 43 ans[o] = MM; 44 } 45 } 46 while(q--) { 47 int x; 48 scanf("%d", &x); 49 printf("%lld\n", ans[x]); 50 } 51 return 0; 52 }
三位数不到1000个，直接暴力
1 #include <bits/stdc++.h> 2 using namespace std; 3 int a[1111]; 4 set<int> S; 5 6 int main() { 7 int n; 8 scanf("%d", &n); 9 for(int i = 1; i <= n; ++i) scanf("%d", a + i); 10 for(int l = 1; l <= 4; ++l) { 11 for(int st = 1; st + l - 1 <= n; ++st) { 12 int t = 0; 13 for(int i = st; i <= st + l - 1; ++i) t = t * 10 + a[i]; 14 S.insert(t); 15 } 16 } 17 for(int i = 0; ; ++i) { 18 if(S.find(i) == S.end()) { 19 printf("%d\n", i); 20 break; 21 } 22 } 23 return 0; 24 }
BFS
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n, cnt[1111][1111], vis[1111][1111]; 4 int dx[] = {0, 1, 0, -1}; 5 int dy[] = {-1, 0, 1, 0}; 6 7 typedef pair<int, int> pii; 8 9 10 bool in(int i, int j) { 11 return i >= 1 && i <= n && j >= 1 && j <= n; 12 } 13 14 int main() { 15 int m; 16 scanf("%d %d", &n, &m); 17 int tot = m; 18 queue<pii> q; 19 for(int i = 1; i <= m; ++i) { 20 int x, y; 21 scanf("%d %d", &x, &y); 22 q.push(pii(x, y)); 23 vis[x][y] = 1; 24 } 25 while(!q.empty()) { 26 pii tmp = q.front(); q.pop(); 27 int x = tmp.first, y = tmp.second; 28 for(int d = 0; d < 4; ++d) { 29 int nx = x + dx[d], ny = y + dy[d]; 30 if(!in(nx, ny)) continue; 31 if(vis[nx][ny]) continue; 32 cnt[nx][ny]++; 33 if(cnt[nx][ny] >= 2) { 34 q.push(pii(nx, ny)); 35 tot++; 36 vis[nx][ny] = 1; 37 } 38 } 39 } 40 puts(tot == n * n ? "YES" : "NO"); 41 return 0; 42 }
题意感人害我中奖，Floyd状压
1 #include <bits/stdc++.h> 2 using namespace std; 3 int G[222][222], id[22], f[1<<15][16]; 4 const int INF = 1e9; 5 6 int main() { 7 int n, m, R; 8 scanf("%d %d %d", &n, &m, &R); 9 for(int i = 1; i <= R; ++i) scanf("%d", id + i); 10 for(int i = 1; i <= n; ++i) 11 for(int j = i + 1; j <= n; ++j) 12 G[i][j] = G[j][i] = INF; 13 for(int i = 1; i <= m; ++i) { 14 int A, B, C; 15 scanf("%d %d %d", &A, &B, &C); 16 G[A][B] = G[B][A] = C; 17 } 18 for(int k = 1; k <= n; ++k) 19 for(int i = 1; i <= n; ++i) 20 for(int j = 1; j <= n; ++j) 21 G[i][j] = min(G[i][j], G[i][k] + G[k][j]); 22 for(int i = 0; i < (1 << R); ++i) 23 for(int j = 1; j <= R; ++j) 24 f[i][j] = INF; 25 for(int i = 1; i <= R; ++i) f[1<<(i - 1)][i] = 0; 26 for(int i = 0; i < (1 << R); ++i){ 27 for(int j = 1; j <= R; ++j) { 28 if(f[i][j] == INF) continue; 29 for(int k = 1; k <= R; ++k) { 30 if(i & (1 << (k - 1))) continue; 31 f[i ^ (1 << (k - 1))][k] = min(f[i ^ (1 << (k - 1))][k], f[i][j] + G[id[j]][id[k]]); 32 } 33 } 34 } 35 int ans = INF; 36 for(int i = 1; i <= R; ++i) ans = min(ans, f[(1<<R)-1][i]); 37 printf("%d\n", ans); 38 return 0; 39 }
直接字符串
1 #include <bits/stdc++.h> 2 using namespace std; 3 char s[1111111]; 4 5 int main() { 6 cin >> s; 7 cout << s << endl; 8 return 0; 9 }
把操作2看成对B串增加111或者000，最后只要判一下模3余数
1 #include <bits/stdc++.h> 2 using namespace std; 3 char a[111], b[111]; 4 5 int main() { 6 int T; 7 scanf("%d", &T); 8 while(T--) { 9 int n, m, x = 0, y = 0; 10 scanf("%d %d %s %s", &n, &m, a + 1, b + 1); 11 for(int i = 1; i <= n; ++i) 12 if(a[i] == '0') y = (y + 1) % 3; 13 else y = (y + 2) % 3; 14 for(int i = 1; i <= m; ++i) { 15 if(b[i] - '0' != x) x ^= 1, y = y * 2 % 3; 16 y = (y + 2) % 3; 17 } 18 puts(y ? "NO" : "YES"); 19 } 20 return 0; 21 }
很好写的模拟？
1 #include <bits/stdc++.h> 2 using namespace std; 3 int dx[] = {-1, 0, 1, -1, 1, -1, 0, 1}; 4 int dy[] = {-1, -1, -1, 0, 0, 1, 1, 1}; 5 int G[11][11]; 6 7 int flip(int x, int y, int o) { 8 int ret = 0; 9 for(int d = 0; d < 8; ++d) { 10 int nx = x + dx[d], ny = y + dy[d]; 11 if(G[nx][ny] != 1 - o) continue; 12 for(int xx = nx + dx[d], yy = ny + dy[d]; ; xx += dx[d], yy += dy[d]) { 13 if(G[xx][yy] == -1) break; 14 if(G[xx][yy] == o) { 15 for(int tx = nx, ty = ny; tx != xx || ty != yy; tx += dx[d], ty += dy[d]) { 16 G[tx][ty] ^= 1; 17 ret++; 18 } 19 break; 20 } 21 } 22 } 23 return ret; 24 } 25 26 int main() { 27 memset(G, -1, sizeof(G)); 28 G[4][5] = G[5][4] = 1; 29 G[4][4] = G[5][5] = 0; 30 int x, y, o = 1; 31 while(~scanf("%d %d", &x, &y)) { 32 if(!flip(x, y, o)) flip(x, y, o ^ 1), G[x][y] = o ^ 1; 33 else G[x][y] = o, o ^= 1; 34 } 35 int w = 0, b = 0; 36 for(int i = 1; i <= 8; ++i) 37 for(int j = 1; j <= 8; ++j) 38 if(G[i][j] == 1) b++; 39 else if(G[i][j] == 0) w++; 40 printf("%d:%d\n", b, w); 41 return 0; 42 }
贪心从后往前吃，可以用一个栈维护没吃完的
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 1e6 + 10; 5 stack<int> st; 6 LL a[maxn]; 7 8 int main() { 9 int n, m, x; 10 scanf("%d %d %d", &n, &m, &x); 11 for(int i = 1; i <= n; ++i) scanf("%lld", a + i); 12 LL sum = 0, ans = 0; 13 for(int i = 1; i < m; ++i) st.push(i), sum += a[i]; 14 for(int i = m; i <= n; ++i) { 15 sum -= a[i-m]; 16 sum += a[i]; 17 st.push(i); 18 while(sum > x) { 19 int id = st.top(); 20 LL y = min(sum - x, a[id]); 21 a[id] -= y; 22 sum -= y; 23 ans += y; 24 if(a[id] == 0) st.pop(); 25 } 26 } 27 printf("%lld\n", ans); 28 return 0; 29 }
枚举一下连接点
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 int L[maxn], R[maxn]; 5 char s[maxn]; 6 7 int main() { 8 int n, tot = 0; 9 scanf("%d %s", &n, s + 1); 10 for(int i = 1; i <= n; ++i) { 11 if(s[i] == '0') L[i] = L[i-1] + 1, tot++; 12 else L[i] = 0; 13 } 14 for(int i = n; i >= 1; --i) { 15 if(s[i] == '0') R[i] = R[i+1] + 1; 16 else R[i] = 0; 17 } 18 int ans = 0; 19 for(int i = 1; i <= n; ++i) { 20 if(L[i] == tot) ans = tot; 21 else ans = max(ans, L[i] + 1); 22 } 23 for(int i = 2; i < n; ++i) { 24 int x = L[i-1] + R[i+1]; 25 if(x != tot) ans = max(ans, x + 1); 26 } 27 printf("%d\n", ans); 28 return 0; 29 }
每次把目前最大的数加到一个重复的数上
1 #include <bits/stdc++.h> 2 using namespace std; 3 set<int> S; 4 5 int main() { 6 int n, x, ans = 0; 7 scanf("%d", &n); 8 for(int i = 1; i <= n; ++i) { 9 scanf("%d", &x); 10 if(S.find(x) != S.end()) ans++; 11 S.insert(x); 12 } 13 printf("%d\n", ans); 14 return 0; 15 }
