打得还可以，10题，最后没能做出B有点可惜。
由于刚刚打完，大部分的题目都是我翻译完之后队友做的，故只贴出代码。
后面的题目很简单，B和E比较难。
A：
solver：zyh、lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 vector<int> li[100001]; 4 long long pTotal[100001]; 5 long long Size[100001]; 6 void dfs(int x, int f, int d, int root) { 7 Size[x] = 1; 8 pTotal[root] += d; 9 for (auto i : li[x]) { 10 if (i == f) continue; 11 dfs(i, x, d + 1, root); 12 Size[x] += Size[i]; 13 } 14 } 15 long long Min = 1e18; 16 void calpTotal(int x, int f, int sum) { 17 if (x != f) pTotal[x] = pTotal[f] + sum - 2 * Size[x]; 18 Min = min(Min, pTotal[x]); 19 for (auto i : li[x]) { 20 if (i == f) continue; 21 calpTotal(i, x, sum); 22 } 23 } 24 void getInnerSum(int x, int f, long long &innerSum, int sum) { 25 for (auto i : li[x]) { 26 if (i == f) continue; 27 innerSum += Size[i] * (sum - Size[i]); 28 getInnerSum(i, x, innerSum, sum); 29 } 30 } 31 int main() { 32 int n; 33 scanf("%d", &n); 34 for (int i = 0; i < n - 2; ++i) { 35 int u, v; 36 scanf("%d%d", &u, &v); 37 li[u].push_back(v); 38 li[v].push_back(u); 39 } 40 int root1, root2; 41 for (int i = 1; i <= n; ++i) 42 if (!Size[i]) { 43 root1 = i; 44 dfs(i, i, 0, i); 45 break; 46 } 47 for (int i = 1; i <= n; ++i) 48 if (!Size[i]) { 49 root2 = i; 50 dfs(i, i, 0, i); 51 break; 52 } 53 Min = 1e18; 54 calpTotal(root1, root1, Size[root1]); 55 long long totalA = Min; 56 57 Min = 1e18; 58 calpTotal(root2, root2, Size[root2]); 59 long long totalB = Min; 60 61 totalA *= Size[root2]; 62 totalB *= Size[root1]; 63 64 long long innerSum1 = 0, innerSum2 = 0; 65 getInnerSum(root1, root1, innerSum1, Size[root1]); 66 getInnerSum(root2, root2, innerSum2, Size[root2]); 67 //cout<<root1<<' '<<root2<<endl; 68 //cout<<Size[root1]*Size[root2]<<' '<<totalA<<' '<<totalB<<' '<<innerSum1<<' '<<innerSum2<<endl; 69 printf("%lld ", Size[root1]*Size[root2] + totalA + totalB + innerSum1 + innerSum2); 70 }
B：
很恶心的概率dp。
C：
solver：czq
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define pb emplace_back 6 #define mp make_pair 7 #define sot(a,mul) sort(a+1,a+1+mul) 8 #define eps 1e-8 9 #define int_inf 0x3f3f3f3f 10 #define ll_inf 0x7f7f7f7f7f7f7f7f 11 #define lson (curnumber<<1) 12 #define rson (curnumber<<1|1) 13 /* namespace */ 14 using namespace std; 15 /* header end */ 16 17 const int maxn = 1e5 + 10, segLim = 10000; 18 vector<int>a, mul, huzhi[maxn]; 19 int n, cnt[maxn]; 20 ll ans = 0; 21 22 void init() { 23 ans = 0; a.clear(); mul.clear(); 24 for (int i = 0; i < maxn; i++) { 25 cnt[i] = 0; huzhi[i].clear(); 26 } 27 } 28 29 int main() { 30 while (~scanf("%d", &n)) { 31 init(); 32 for (int i = 0; i < n; i++) { 33 int x; scanf("%d", &x); 34 cnt[x]++; a.pb(x); 35 } 36 sort(a.begin(), a.end()); 37 a.erase(unique(a.begin(), a.end()), a.end()); 38 for (int i = 0; i <= segLim; i++) if (cnt[i] >= 2) mul.pb(i); 39 for (int i = 1; i <= segLim; i++) 40 for (int j = 0; j < (int)mul.size(); j++) 41 if (__gcd(i, mul[j]) == 1) huzhi[i].pb(mul[j]); 42 // n^2, enum shang di && xia di 43 for (int i = 0; i < (int)a.size(); i++) { 44 for (int j = i + 1; j < (int)a.size(); j++) { 45 int gcd = __gcd(a[i], a[j]); 46 int curr = 0; 47 if ((a[j] - a[i]) % 2 == 0) curr = (a[j] - a[i]) / 2 + 1; 48 else curr = (a[j] - a[i] + 1) / 2; 49 int number = lower_bound(huzhi[gcd].begin(), huzhi[gcd].end(), curr) - huzhi[gcd].begin(); 50 ans += (int)huzhi[gcd].size() - number; 51 if (gcd == 1) { 52 if (a[i] >= curr && cnt[a[i]] == 2) ans--; 53 if (a[j] >= curr && cnt[a[j]] == 2) ans--; 54 } 55 } 56 } 57 printf("%lld ", ans); 58 } 59 return 0; 60 }
D：
solver：zyh
1 #include <iostream> 2 using namespace std; 3 int a[500001]; 4 int Max[1001]; 5 int bin[1001]; 6 int main() { 7 int n, c; 8 scanf("%d%d", &n, &c); 9 for (int i = 0; i < n; ++i) { 10 scanf("%d", &a[i]); 11 } 12 int ans = 0; 13 for (int j = 1; j <= c; ++j) { 14 int cnt = 0; 15 for (int i = 0; i < n; ++i) { 16 if (a[i] == j) ++cnt; 17 else { 18 if (Max[a[i]] < cnt) { 19 bin[a[i]]++; 20 Max[a[i]] = cnt; 21 } 22 } 23 } 24 //cout<<"done"; 25 for (int i = 1; i <= c; ++i) { 26 int s = bin[i] * 2; 27 if (s == 0) continue; 28 if (Max[i] < cnt) ++s; 29 ans = max(ans, s); 30 bin[i] = 0; 31 Max[i] = 0; 32 } 33 //cout<<"done2"; 34 } 35 printf("%d ", ans); 36 }
E：
solver：zyh、lzh
细节恶心的模拟题。
1 #include <iostream> 2 #include <algorithm> 3 #include <queue> 4 using namespace std; 5 6 struct item { 7 int t, x; 8 item() {} 9 item(int _t, int _x) { 10 t = _t; 11 x = _x; 12 } 13 bool operator< (const item &b)const { 14 return t < b.t; 15 } 16 }; 17 struct segment { 18 int l, r; 19 segment() {} 20 segment(int _l, int _r) { 21 l = _l; 22 r = _r; 23 } 24 bool operator<(const segment &b)const { 25 if (l != b.l) return l < b.l; 26 return r < b.r; 27 } 28 }; 29 item items[1000001]; 30 int len[1000001]; 31 vector<int> buffers[1000001]; 32 int front[1000001]; 33 vector<segment> overflow; 34 int main() { 35 int n, m; 36 scanf("%d%d", &n, &m); 37 for (int i = 0; i < n; ++i) { 38 int t, x; 39 scanf("%d%d", &t, &x); 40 items[i] = item(t, x); 41 } 42 for (int i = 1; i <= m; ++i) scanf("%d", &len[i]); 43 sort(items, items + n); 44 overflow.clear(); 45 for (int i = 0; i < n; ++i) { 46 buffers[items[i].x].push_back(items[i].t); 47 //cout<<buffers[items[i].x].size()<<endl; 48 if (buffers[items[i].x].size() > len[items[i].x]) { 49 auto t = buffers[items[i].x][front[i]]; 50 front[i]++; 51 //buffers[items[i].x].pop(); 52 overflow.push_back(segment(t, items[i].t)); 53 } 54 } 55 long long ans = 0; 56 int last = 0; 57 for (auto i : overflow) { 58 //cout<<"pp:"<<i.l<<' '<<i.r<<endl; 59 if (i.l > last) last = i.l + 1; 60 else last++; 61 ans += max(0, last - i.r - 1); 62 } 63 printf("%lld ", ans); 64 }
F：
solver：zyh
1 #include <bits/stdc++.h> 2 using namespace std; 3 char str[1000]; 4 int main() { 5 int len; 6 long long a, v, _i, n; 7 while (~scanf("%d", &len)) { 8 scanf("%s", str); 9 a = v = _i = n = 0; 10 for (int i = 0; i < len; ++i) { 11 if (str[i] == 'a') a++; 12 if (str[i] == 'v') v++; 13 if (str[i] == 'i') _i++; 14 if (str[i] == 'n') n++; 15 } 16 long long p, q; 17 //cout<<a<<' '<<v<<' '<<_i<<" "<<n<<endl; 18 p = a * v * _i * n; 19 q = len * len * len * len; 20 long long t = __gcd(p, q); 21 p /= t; q /= t; 22 printf("%lld/%lld ", p, q); 23 } 24 25 }
G：
solver：lzh
这题我还读错了题意害队友wa了两发……
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 #define ff first 5 #define ss second 6 7 int a[1010], b[1010], vis[3010]; 8 int main() { 9 int n, m; 10 while (~scanf("%d%d", &n, &m)) { 11 for (int i = 1; i <= 1000; i++)vis[i] = 0; 12 for (int i = 1; i <= n; i++)scanf("%d", &a[i]), vis[a[i]]++; 13 for (int i = 1; i <= m; i++)scanf("%d", &b[i]); 14 for (int i = 0; i <= 1010; i++) { 15 int ok = 1; 16 for (int j = 1; j <= m; j++) 17 if (vis[b[j] + i]) { 18 ok = 0; 19 break; 20 } 21 if (ok) { 22 printf("%d ", i); 23 break; 24 } 25 } 26 } 27 }
H：
solver：zyh、czq
1 #include <iostream> 2 using namespace std; 3 const long long M = 1000000007; 4 long long qpow(long long x, long long y) { 5 long long ans = 1; 6 while (y) { 7 if (y & 1) ans = (ans * x) % M; 8 x = x * x % M; 9 y >>= 1; 10 } 11 return ans; 12 } 13 int main() { 14 long long n; 15 scanf("%lld", &n); 16 printf("%lld ", (n * (n + 1) / 2 % M) * (qpow(n * n, M - 2) % M) % M); 17 18 }
I：
solver：lzh
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 #define ff first 5 #define ss second 6 7 int main() { 8 int n; 9 while (~scanf("%d", &n)) { 10 double ans = 0; 11 for (int i = 1; i <= n; i++) { 12 char s[20]; scanf("%s", s + 1); 13 int len = strlen(s + 1); 14 if (s[len] >= '5')ans += 0.001 * ('9' - s[len] + 1); 15 else ans -= 0.001 * (s[len] - '0'); 16 } 17 printf("%.3lf ", ans); 18 } 19 }
J：
solver：zyh、lzh
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<ll, ll> pii; 5 #define ff first 6 #define ss second 7 8 ll lcm(ll a, ll b) { 9 return a / __gcd(a, b) * b; 10 } 11 12 ll a[1010]; 13 int main() { 14 int n; ll m; 15 while (~scanf("%d%lld", &n, &m)) { 16 for (int i = 1; i <= n; i++)scanf("%lld", &a[i]); 17 pii cur = make_pair(1, a[1]); 18 for (int i = 2; i <= n; i++) { 19 ll x = lcm(cur.ss, a[i]); 20 cur.ff = cur.ff * (x / cur.ss) + x / a[i]; 21 cur.ss = x; 22 ll g = __gcd(cur.ff, cur.ss); 23 cur.ff /= g, cur.ss /= g; 24 } 25 if (m % cur.ff) { 26 printf("No "); continue; 27 } 28 ll x = m / cur.ff * cur.ss; 29 ll lcmm = 1; 30 for (int i = 1; i <= n; i++)lcmm = lcm(lcmm, a[i]); 31 if (x % lcmm) { 32 printf("No "); continue; 33 } else { 34 printf("Yes "); 35 printf("%lld", x / a[1]); 36 for (int i = 2; i <= n; i++)printf(" %lld", x / a[i]); 37 printf(" "); 38 } 39 } 40 }
K：
solver：zyh
1 #include <iostream> 2 using namespace std; 3 int main() { 4 int a, b; 5 cin >> a >> b; 6 int x = a + b; 7 int y = a - b; 8 cout << x *y / 4 << endl; 9 }