A 深度学习
puts(n)
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 double n; 7 while (scanf("%lf", &n) != EOF) 8 printf("%.10f ", n); 9 return 0; 10 }
B 异或求和
思路： 有一个$O(nlogn^2)$ 的做法，但是T了。
设$bit(x, i)$为 x在二进制下第i位为1还是0
考虑 $sum_{1 <= i < j < k <= n}(a_i oplus a_j)(a_j oplus a_k)(a_i oplus a_k)$
将最后一部分的 $(a_i oplus a_k)$ 拆分成 $sum_{l = 0}^{l = 29}(bit(a_i, l))(bit(a_k, l))$
那么我们再枚举j 就可以
T了的代码：
#include <bits/stdc++.h> using namespace std; #define N 100010 #define ll long long ll MOD = (ll)998244353; int n; int arr[N], prefix[30][2][30][2], suffix[30][2][30][2]; void Run() { while (scanf("%d", &n) != EOF) { for (int i = 1; i <= n; ++i) scanf("%d", arr + i); bitset <30> b; for (int i = 2; i <= n; ++i) { b = arr[i]; for (int j = 29; j >= 0; --j) for (int k = 29; k >= 0; --k) ++suffix[j][b[j]][k][b[k]]; } b = arr[1]; for (int j = 29; j >= 0; --j) for (int k = 29; k >= 0; --k) ++prefix[j][b[j]][k][b[k]]; ll ans = 0; for (int i = 2; i < n; ++i) { b = arr[i]; for (int j = 29; j >= 0; --j) for (int k = 29; k >= 0; --k) --suffix[j][b[j]][k][b[k]]; for (int j = 29; j >= 0; --j) { ll l = 0, r = 0; // enum 0 1 for (int k = 29; k >= 0; --k) { l = (l + ((1ll << k) * prefix[j][0][k][b[k] ^ 1]) % MOD) % MOD; r = (r + ((1ll << k) * suffix[j][1][k][b[k] ^ 1]) % MOD) % MOD; } ans = (ans + (1ll << j) * l % MOD * r % MOD) % MOD; l = 0, r = 0; // enum 1 0 for (int k = 29; k >= 0; --k) { l = (l + ((1ll << k) * prefix[j][1][k][b[k] ^ 1]) % MOD) % MOD; r = (r + ((1ll << k) * suffix[j][0][k][b[k] ^ 1]) % MOD) % MOD; } ans = (ans + (1ll << j) * l % MOD * r % MOD) % MOD; } for (int j = 29; j >= 0; --j) for (int k = 29; k >= 0; --k) ++prefix[j][b[j]][k][b[k]]; } printf("%lld ", ans); } } int main() { #ifdef LOCAL freopen("Test.in", "r", stdin); #endif Run(); return 0; }
C 异或计数
留坑。
D 最小生成树
思路：用最小的点权去连边
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define ll long long 6 7 int n; 8 9 int main() 10 { 11 while (scanf("%d", &n) != EOF) 12 { 13 ll res = 0, num, Min = 0x3f3f3f3f3f3f3f3f; 14 for (int i = 1; i <= n; ++i) 15 { 16 scanf("%lld ", &num); 17 Min = min(Min, num); 18 res += num; 19 } 20 printf("%lld ", res + Min * (n - 2)); 21 } 22 return 0; 23 }
E 乒乓球
留坑。
F 导数卷积
留坑。
G 区间权值
显然可以发现长度为l和n-l的权值相同，然后枚举每个长度l，发现出现次数为两端向中间增加，然后中间不变，最后O(n)扫一遍即可
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int maxn = (int)3e5 + 10; 8 const ll MOD = 1e9 + 7; 9 int n; 10 ll ans; 11 ll w[maxn]; 12 ll val[maxn]; 13 14 int main() 15 { 16 while(~scanf("%d", &n)) 17 { 18 for(int i = 1; i <= n; ++i) 19 { 20 scanf("%lld", val + i); 21 } 22 for(int i = 1; i <= n; ++i) 23 { 24 scanf("%lld", w + i); 25 } 26 for(int i = 1; i <= n; ++i) 27 { 28 val[i] += val[i - 1]; 29 val[i] %= MOD; 30 } 31 ans = 0; 32 ll tmp = 0; 33 for(int i = 1; i <= n / 2; ++i) 34 { 35 tmp += (val[n - i + 1] - val[i - 1] + MOD) % MOD; 36 tmp %= MOD; 37 ans += (tmp * w[i] % MOD + tmp * w[n - i + 1] % MOD) % MOD; 38 ans %= MOD; 39 } 40 if(n & 1) 41 { 42 tmp += val[(n + 1) / 2] - val[n / 2]; 43 tmp %= MOD; 44 ans += tmp * w[(n + 1) / 2] % MOD; 45 ans %= MOD; 46 } 47 printf("%lld ", ans); 48 } 49 return 0; 50 }
H 树链博弈
思路：每层上都有偶数个黑点，后手必胜，否则，先手必胜
假设所有点都是白点，是先手必败态
如果先手必胜，那么走一步，走到先手必败态
如果是先手必败态，走一步，可以走到先手必胜态
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 6 int n; 7 int color[N], deep[N]; 8 vector <int> G[N]; 9 10 void DFS(int u, int fa, int deepth) 11 { 12 deep[deepth] += color[u]; 13 for (auto v : G[u]) 14 { 15 if (v == fa) continue; 16 DFS(v, u, deepth + 1); 17 } 18 } 19 20 void Run() 21 { 22 while (scanf("%d", &n) != EOF) 23 { 24 for (int i = 1; i <= n; ++i) scanf("%d", color + i); 25 for (int i = 1, u, v; i < n; ++i) 26 { 27 scanf("%d%d", &u, &v); 28 G[u].push_back(v); 29 G[v].push_back(u); 30 } 31 DFS(1, 1, 0); 32 int res = 0; 33 for (int i = 0; i < n; ++i) if (deep[i] & 1) res = 1; 34 puts(res ? "First" : "Second"); 35 } 36 } 37 38 int main() 39 { 40 #ifdef LOCAL 41 freopen("Test.in", "r", stdin); 42 #endif 43 44 Run(); 45 return 0; 46 }
I 连通块计数
分两种情况考虑，一种是含根，一种是不含根，分别计数
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() { 6 long long ans, t ,x, i, n, MOD; 7 scanf("%lld ",&n); 8 MOD = 998244353; 9 ans = 0; t = 1; 10 for (i = 1; i <= n; ++i) { 11 scanf("%lld ",&x); 12 ans = (ans + (x * x + x) / 2) % MOD; 13 t = (t * (x + 1)) % MOD; 14 } 15 ans = (ans + t) % MOD; 16 printf("%lld ",ans); 17 return 0; 18 }
J 寻找复读机
按题意模拟即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 6 int n, m; 7 int used[N], vis[N]; 8 vector <int> ans; 9 char s[N][110]; 10 11 int main() 12 { 13 while (scanf("%d%d", &n, &m) != EOF) 14 { 15 memset(vis, 0, sizeof vis); 16 memset(used, 0, sizeof used); 17 ans.clear(); 18 for (int i = 1, x; i <= m; ++i) 19 { 20 scanf("%d %s", &x, s[i]); 21 if (i == 1) vis[x] = 1; 22 else if (strcmp(s[i], s[i - 1]) != 0) vis[x] = 1; 23 used[x] = 1; 24 } 25 for (int i = 1; i <= n; ++i) if (!vis[i]) 26 ans.push_back(i); 27 if (!ans.size()) puts(""); 28 for (int i = 0, len = ans.size(); i < len; ++i) printf("%d%c", ans[i], " "[i == len - 1]); 29 } 30 return 0; 31 }