A - Airport Bus
贪心，能取就取。
1 //waz 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 #define mp make_pair 7 #define pb push_back 8 #define fi first 9 #define se second 10 #define ALL(x) (x).begin(), (x).end() 11 #define SZ(x) ((int)((x).size())) 12 13 typedef pair<int, int> PII; 14 typedef vector<int> VI; 15 typedef long long int64; 16 typedef unsigned int uint; 17 typedef unsigned long long uint64; 18 19 #define gi(x) ((x) = F()) 20 #define gii(x, y) (gi(x), gi(y)) 21 #define giii(x, y, z) (gii(x, y), gi(z)) 22 23 int F() 24 { 25 char ch; 26 int x, a; 27 while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); 28 if (ch == '-') ch = getchar(), a = -1; 29 else a = 1; 30 x = ch - '0'; 31 while (ch = getchar(), ch >= '0' && ch <= '9') 32 x = (x << 1) + (x << 3) + ch - '0'; 33 return a * x; 34 } 35 36 const int N = 1e5 + 10; 37 38 int n, c, k, t[N], ans; 39 40 int main() 41 { 42 giii(n, c, k); 43 for (int i = 1; i <= n; ++i) gi(t[i]); 44 sort(t + 1, t + n + 1); 45 for (int i = 1; i <= n; ++i) 46 { 47 int j; 48 if (t[n] > t[i] + k) j = upper_bound(t + 1, t + n + 1, t[i] + k) - t - 1; 49 else j = n; 50 j = min(i + c - 1, j); 51 i = j; 52 ++ans; 53 } 54 printf("%d\n", ans); 55 }
B - Colorful Creatures
二分答案，直接判断即可。
1 //waz 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 #define mp make_pair 7 #define pb push_back 8 #define fi first 9 #define se second 10 #define ALL(x) (x).begin(), (x).end() 11 #define SZ(x) ((int)((x).size())) 12 13 typedef pair<int, int> PII; 14 typedef vector<int> VI; 15 typedef long long int64; 16 typedef unsigned int uint; 17 typedef unsigned long long uint64; 18 19 #define gi(x) ((x) = F()) 20 #define gii(x, y) (gi(x), gi(y)) 21 #define giii(x, y, z) (gii(x, y), gi(z)) 22 23 int F() 24 { 25 char ch; 26 int x, a; 27 while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); 28 if (ch == '-') ch = getchar(), a = -1; 29 else a = 1; 30 x = ch - '0'; 31 while (ch = getchar(), ch >= '0' && ch <= '9') 32 x = (x << 1) + (x << 3) + ch - '0'; 33 return a * x; 34 } 35 36 const int N = 1e5 + 10; 37 38 int n; 39 40 int a[N]; 41 42 bool check(int x) 43 { 44 int64 s = a[x]; 45 for (int j = 1; j <= n; ++j) 46 if (x != j) 47 { 48 if (s * 2 >= a[j]) s += a[j]; 49 else return 0; 50 } 51 return 1; 52 }
C - Squared Graph
把原图中孤立点、二分图和其它连通块分开考虑一下即可。
1 //waz 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 #define mp make_pair 7 #define pb push_back 8 #define fi first 9 #define se second 10 #define ALL(x) (x).begin(), (x).end() 11 #define SZ(x) ((int)((x).size())) 12 13 typedef pair<int, int> PII; 14 typedef vector<int> VI; 15 typedef long long int64; 16 typedef unsigned int uint; 17 typedef unsigned long long uint64; 18 19 #define gi(x) ((x) = F()) 20 #define gii(x, y) (gi(x), gi(y)) 21 #define giii(x, y, z) (gii(x, y), gi(z)) 22 23 int F() 24 { 25 char ch; 26 int x, a; 27 while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); 28 if (ch == '-') ch = getchar(), a = -1; 29 else a = 1; 30 x = ch - '0'; 31 while (ch = getchar(), ch >= '0' && ch <= '9') 32 x = (x << 1) + (x << 3) + ch - '0'; 33 return a * x; 34 } 35 36 const int N = 1e5 + 10, M = 1e5 + 10; 37 38 int n, m; 39 40 bool flag; 41 42 VI edge[N]; 43 44 int co[N]; 45 46 int O, P, Q; 47 48 void dfs(int i, int x) 49 { 50 co[i] = x; 51 for (auto j : edge[i]) 52 { 53 if (co[j]) 54 { 55 if (co[j] != (co[i] ^ 1)) flag = 0; 56 } 57 else 58 { 59 dfs(j, x ^ 1); 60 } 61 } 62 } 63 64 int main() 65 { 66 gii(n, m); 67 for (int i = 1; i <= m; ++i) 68 { 69 int u, v; 70 gii(u, v); 71 edge[u].pb(v); 72 edge[v].pb(u); 73 } 74 for (int i = 1; i <= n; ++i) 75 { 76 if (!co[i]) 77 { 78 if (!SZ(edge[i])) 79 { 80 ++O; 81 continue; 82 } 83 flag = 1; 84 dfs(i, 2); 85 if (flag) ++Q; 86 else ++P; 87 } 88 } 89 long long ans = 2LL * O * n - 1LL * O * O + 1LL * P * P + 2LL * P * Q + 2LL * Q * Q; 90 printf("%lld\n", ans); 91 return 0; 92 }
D - Half Reflector
我们发现2*n次之后肯定就是一个循环了。前面的暴力即可。
1 //waz 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 #define mp make_pair 7 #define pb push_back 8 #define fi first 9 #define se second 10 #define ALL(x) (x).begin(), (x).end() 11 #define SZ(x) ((int)((x).size())) 12 13 typedef pair<int, int> PII; 14 typedef vector<int> VI; 15 typedef long long int64; 16 typedef unsigned int uint; 17 typedef unsigned long long uint64; 18 19 #define gi(x) ((x) = F()) 20 #define gii(x, y) (gi(x), gi(y)) 21 #define giii(x, y, z) (gii(x, y), gi(z)) 22 23 int F() 24 { 25 char ch; 26 int x, a; 27 while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); 28 if (ch == '-') ch = getchar(), a = -1; 29 else a = 1; 30 x = ch - '0'; 31 while (ch = getchar(), ch >= '0' && ch <= '9') 32 x = (x << 1) + (x << 3) + ch - '0'; 33 return a * x; 34 } 35 36 const int N = 2e5 + 10; 37 38 char str[N]; 39 40 int n, k; 41 42 int pos(int x, int l) 43 { 44 int p = x + l % n; 45 if (p > n) p -= n; 46 return p; 47 } 48 49 int main() 50 { 51 gii(n, k); 52 scanf("%s", str + 1); 53 for (int i = 1; i <= n; ++i) str[i] -= 'A'; 54 if (k <= 2 * n + 10) 55 { 56 int l = 0, t = 0; 57 for (int i = 1; i <= k; ++i) 58 { 59 if (!(str[pos(1, l)] ^ t)) 60 { 61 str[pos(1, l)] ^= 1; 62 } 63 else 64 { 65 ++l, t ^= 1; str[pos(n, l)] = t; 66 } 67 } 68 for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A'); 69 puts(""); 70 return 0; 71 } 72 else 73 { 74 int l = 0, t = 0; 75 for (int i = 1; i <= 2 * n + 10; ++i) 76 { 77 if (!(str[pos(1, l)] ^ t)) 78 { 79 str[pos(1, l)] ^= 1; 80 } 81 else 82 { 83 ++l, t ^= 1; str[pos(n, l)] = t; 84 } 85 } 86 k -= 2 * n + 10; 87 if (n & 1) 88 { 89 str[pos(1, l)] ^= (k & 1); 90 for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A'); 91 puts(""); 92 return 0; 93 } 94 else 95 { 96 for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A'); 97 puts(""); 98 return 0; 99 } 100 } 101 }
E - Increasing Numbers
我们发现一个上升的数肯定可以拆成很多个不同位数的111111……之和。就是(10^x-1)/9，我们暴力加一下，直接判断即可。
1 //waz 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 #define mp make_pair 7 #define pb push_back 8 #define fi first 9 #define se second 10 #define ALL(x) (x).begin(), (x).end() 11 #define SZ(x) ((int)((x).size())) 12 13 typedef pair<int, int> PII; 14 typedef vector<int> VI; 15 typedef long long int64; 16 typedef unsigned int uint; 17 typedef unsigned long long uint64; 18 19 #define gi(x) ((x) = F()) 20 #define gii(x, y) (gi(x), gi(y)) 21 #define giii(x, y, z) (gii(x, y), gi(z)) 22 23 int F() 24 { 25 char ch; 26 int x, a; 27 while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); 28 if (ch == '-') ch = getchar(), a = -1; 29 else a = 1; 30 x = ch - '0'; 31 while (ch = getchar(), ch >= '0' && ch <= '9') 32 x = (x << 1) + (x << 3) + ch - '0'; 33 return a * x; 34 } 35 36 int n; 37 38 char s[500010]; 39 40 int x[500010]; 41 42 int sum; 43 44 int main() 45 { 46 scanf("%s", s + 1); 47 n = strlen(s + 1); 48 reverse(s + 1, s + n + 1); 49 for (int i = 1; i <= n; ++i) x[i] = (s[i] - 48) * 9; 50 for (int i = 1; i <= n; ++i) x[i + 1] += x[i] / 10, x[i] %= 10; 51 if (x[n + 1]) ++n; 52 for (int i = 1; i <= n; ++i) sum += x[i]; 53 for (int k = 1; k <= 2 * n; ++k) 54 { 55 int v = 9, j = 1; 56 while (v) 57 { 58 sum -= x[j]; 59 x[j] += v; 60 v = x[j] / 10; 61 x[j] %= 10; 62 sum += x[j]; 63 if (j > n) ++n; 64 ++j; 65 } 66 if (sum <= 9 * k) 67 { 68 printf("%d\n", k); 69 return 0; 70 } 71 } 72 return 0; 73 }
F - Train Service Planning
题目都不是很懂（太菜了qwq）。。。先咕着吧。