A Live Love
水。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 const double eps = 1e-8; 7 const int INF = 0x3f3f3f3f; 8 const ll MOD = (int)1e9 + 7; 9 const int maxn = (int)1e6 + 10; 10 11 int n, m; 12 13 inline void RUN() 14 { 15 int t; 16 scanf("%d", &t); 17 while (t--) 18 { 19 scanf("%d %d", &n, &m); 20 int ans1 = m; 21 int ans2 = (n) / (n - m + 1); 22 printf("%d %d\n", ans1, ans2); 23 } 24 } 25 26 int main() 27 { 28 #ifdef LOCAL_JUDGE 29 freopen("Text.txt", "r", stdin); 30 #endif // LOCAL_JUDGE 31 32 RUN(); 33 34 #ifdef LOCAL_JUDGE 35 fclose(stdin); 36 #endif // LOCAL_JUDGE 37 return 0; 38 }
B Red Black Tree
题意：有一个树，上面有红点和黑点，有边权，每个点的花费定义为它到离它最近的红点的距离，每次询问给出一些点，能够将这棵树中的一个黑点变为红点，使得这些点中的最大花费最小
思路：二分答案，符合条件的点不管，将不符合条件的点LCA求出来，变红，然后算距离
C Halting Problem
题意：
思路：
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 const double eps = 1e-8; 7 const int INF = 0x3f3f3f3f; 8 const ll MOD = (int)1e9 + 7; 9 const int maxn = (int)1e4 + 10; 10 11 int n, num; 12 int vis[maxn][260]; 13 struct node { 14 char op[10]; 15 int v, k; 16 }arr[maxn]; 17 18 inline void RUN() 19 { 20 int t; 21 scanf("%d", &t); 22 while (t--) 23 { 24 scanf("%d", &n); 25 for (int i = 1; i <= n; ++i) 26 { 27 for (int j = 0; j < 256; ++j) 28 { 29 vis[i][j] = 0; 30 } 31 } 32 int flag = true; 33 int s = 0; 34 int now = 1; 35 for (int i = 1; i <= n; ++i) 36 { 37 scanf("%s", arr[i].op); 38 if (strcmp(arr[i].op,"add") == 0) 39 { 40 scanf("%d", &arr[i].v); 41 arr[i].k = 0; 42 } 43 else 44 { 45 scanf("%d %d", &arr[i].v, &arr[i].k); 46 } 47 } 48 while (1) 49 { 50 if (now == n + 1) 51 { 52 flag = true; 53 break; 54 } 55 if (vis[now][s]) 56 { 57 flag = false; 58 break; 59 } 60 vis[now][s]++; 61 if (strcmp(arr[now].op, "add") == 0) 62 { 63 s = (s + arr[now].v); 64 if (s >= 256)s -= 256; 65 now++; 66 } 67 else if (strcmp(arr[now].op, "beq") == 0) 68 { 69 if (s == arr[now].v) 70 { 71 now = arr[now].k; 72 } 73 else 74 { 75 now++; 76 } 77 } 78 else if (strcmp(arr[now].op, "bne") == 0) 79 { 80 if (s != arr[now].v) 81 { 82 now = arr[now].k; 83 } 84 else 85 { 86 now++; 87 } 88 } 89 else if (strcmp(arr[now].op, "blt") == 0) 90 { 91 if (s < arr[now].v) 92 { 93 now = arr[now].k; 94 } 95 else 96 { 97 now++; 98 } 99 } 100 else if (strcmp(arr[now].op, "bgt") == 0) 101 { 102 if (s > arr[now].v) 103 { 104 now = arr[now].k; 105 } 106 else 107 { 108 now++; 109 } 110 } 111 } 112 puts(flag ? "Yes" : "No"); 113 } 114 } 115 116 int main() 117 { 118 #ifdef LOCAL_JUDGE 119 freopen("Text.txt", "r", stdin); 120 #endif // LOCAL_JUDGE 121 122 RUN(); 123 124 #ifdef LOCAL_JUDGE 125 fclose(stdin); 126 #endif // LOCAL_JUDGE 127 return 0; 128 }
D Pixel Art
留坑。
E Infinite Parenthesis Sequence
留坑。
F Chaleur
留坑。
G Couleur
留坑。
H Traveling on the Axis
题意：一个字符串表示当前点是红灯还是绿灯，灯一秒变一次，求$\sum_{p = 0}^{n - 1}\sum_{q = p +1}^{n} t(p, q)$
思路：前缀搞一搞
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 char st[100000+10]; 6 long long f[100000+10]; 7 int main() { 8 long long t,n,i,j,k; 9 long long ans; 10 scanf("%lld",&t); 11 while (t--) { 12 scanf("%s",st); 13 n=strlen(st); 14 f[1]=1; ans=0; 15 for (i=1;i<n;++i) 16 if (st[i]==st[i-1]) f[i+1]=2; 17 else f[i+1]=1; 18 for (i=1;i<=n;++i) { 19 ans+=(f[i])*i*(n-i+1); 20 if (st[i-1]=='0') ans+=n-i+1; 21 if (f[i]==2) ans-=n-i+1; 22 } 23 printf("%lld\n",ans); 24 } 25 return 0; 26 }
J Press the Button
题意：周期性按灯，每按一次，如果灯本来是亮着就++， 求最后+了多少次
思路：
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 const double eps = 1e-8; 7 const int INF = 0x3f3f3f3f; 8 const ll MOD = (int)1e9 + 7; 9 const int maxn = (int)1e4 + 10; 10 11 inline ll gcd(ll a, ll b) 12 { 13 return b == 0 ? a : gcd(b, a%b); 14 } 15 16 ll a, b, c, d, v, t; 17 18 inline void RUN() 19 { 20 int cas; 21 scanf("%d", &cas); 22 while (cas--) 23 { 24 scanf("%lld %lld %lld %lld %lld %lld", &a, &b, &c, &d, &v, &t); 25 ll G = gcd(a, c); 26 G = a * c / G; 27 //cout << G << endl; 28 ll res =0; 29 ll res1 = b - 1 + d; 30 ll len = t / G; 31 ll M = t % G; 32 ll t1 = a; 33 ll t2 = c; 34 ll now = 0; 35 while (t1 <= G && t2 <= G) 36 { 37 if (t1 < t2) 38 { 39 ll tmp = t1 - now; 40 if (tmp > v) 41 { 42 res += b - 1; 43 if (t1 <= M) 44 { 45 res1 += b - 1; 46 } 47 } 48 else 49 { 50 res += b; 51 if (t1 <= M) 52 { 53 res1 += b; 54 } 55 } 56 now = t1; 57 t1 += a; 58 59 } 60 else if (t1 > t2) 61 { 62 ll tmp = t2 - now; 63 if (tmp > v) 64 { 65 res += d - 1; 66 if (t2 <= M) 67 { 68 res1 += d - 1; 69 } 70 } 71 else 72 { 73 res += d; 74 if (t2 <= M) 75 { 76 res1 += d; 77 } 78 } 79 now = t2; 80 t2 += c; 81 } 82 else if (t1 == t2) 83 { 84 ll tmp = t1 - now; 85 if (tmp > v) 86 { 87 res += b + d - 1; 88 if (t1 <= M) 89 { 90 res1 += b + d - 1; 91 } 92 } 93 else 94 { 95 res += b + d; 96 if (t1 <= M) 97 { 98 res1 += b + d; 99 } 100 } 101 now = t1; 102 t1 += a; 103 t2 += c; 104 105 } 106 } 107 ll ans = res * len + res1; 108 printf("%lld\n", ans); 109 } 110 } 111 112 int main() 113 { 114 #ifdef LOCAL_JUDGE 115 freopen("Text.txt", "r", stdin); 116 #endif // LOCAL_JUDGE 117 118 RUN(); 119 120 #ifdef LOCAL_JUDGE 121 fclose(stdin); 122 #endif // LOCAL_JUDGE 123 return 0; 124 }
K XOR Clique
题意：给出n个数，求选出一个点集，使得里面的点两两异或不超过这两个数中小的那个，求点集里面元素个数最大是多少
思路：枚举最高位
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 const double eps = 1e-8; 7 const int INF = 0x3f3f3f3f; 8 const ll MOD = (int)1e9 + 7; 9 const int maxn = (int)1e6 + 10; 10 11 int n, num; 12 int arr[40]; 13 14 inline void RUN() 15 { 16 int t; 17 scanf("%d", &t); 18 while (t--) 19 { 20 int ans = 0; 21 scanf("%d", &n); 22 memset(arr, 0, sizeof arr); 23 for (int i = 1; i <= n; ++i) 24 { 25 scanf("%d", &num); 26 int cnt = 0; 27 while (num) 28 { 29 cnt++; 30 num >>= 1; 31 } 32 arr[cnt]++; 33 ans = max(ans, arr[cnt]); 34 } 35 printf("%d\n", ans); 36 } 37 } 38 39 int main() 40 { 41 #ifdef LOCAL_JUDGE 42 freopen("Text.txt", "r", stdin); 43 #endif // LOCAL_JUDGE 44 45 RUN(); 46 47 #ifdef LOCAL_JUDGE 48 fclose(stdin); 49 #endif // LOCAL_JUDGE 50 return 0; 51 }