比较简单的三星场，差一点AK
最近忙着面试，三个人都不在状态，我前期很C后期萎了，两个队友前期很萎后面很C，可能这就是队内文化8
题目链接：https://codeforces.com/gym/101858
A：
solver：zyh、lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 long long dp[2][8]; 4 const long long M = 1e9 + 7; 5 int main() { 6 int n; 7 scanf("%d", &n); 8 dp[1][0] = dp[1][6] = dp[1][3] = 1; 9 for (int i = 2; i <= n; ++i) { 10 int now = i & 1; 11 int pre = (i - 1) & 1; 12 dp[now][7] = (dp[pre][0] + dp[pre][6] + dp[pre][3]) % M; 13 dp[now][6] = (dp[pre][1] + dp[pre][7]) % M; 14 dp[now][5] = dp[pre][2]; 15 dp[now][4] = dp[pre][3]; 16 dp[now][3] = (dp[pre][4] + dp[pre][7]) % M; 17 dp[now][2] = dp[pre][5]; 18 dp[now][1] = dp[pre][6]; 19 dp[now][0] = dp[pre][7]; 20 //cout<<dp[now][7]<<endl; 21 } 22 printf("%lld", dp[n & 1][7]); 23 }
B：
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 eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 string s, ans = ""; 15 int numC = 0, numS = 0; 16 17 int main() { 18 ios::sync_with_stdio(false); 19 cin.tie(0); 20 cin >> s; 21 for (auto ch : s) { 22 if (ch == 'C' && numC <= 1) { 23 numC++, numS = 0; 24 ans += 'B'; 25 } else if (ch == 'S' && numS <= 1) { 26 numS++, numC = 0; 27 ans += 'D'; 28 } else if (ch == 'C') { 29 numC = 0, ans += 'P'; 30 } else if (ch == 'S') { 31 numS = 0, ans += 'T'; 32 } 33 } 34 cout << ans << endl; 35 return 0; 36 }
C：
solver：lzh
据说这个题随便剪一下枝就能过了，跑得飞起
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<int, int> pii; 5 #define ff first 6 #define ss second 7 8 int a[10][10], ok = 0; 9 int w[10][10], quyu[40], num[40]; 10 int rvis[10][10], cvis[10][10], blockvis[10][10], quyuvis[40][10]; 11 inline int getblock(int x, int y) { 12 return (x + 1) / 2 * 2 - (y > 3 ? 0 : 1); 13 } 14 void dfs(int x, int y) { 15 if (y == 7) { 16 x++, y = 1; 17 dfs(x, y); 18 return; 19 } 20 if (ok) 21 return; 22 if (x == 7) { 23 for (int i = 1; i <= 6; i++) 24 for (int j = 1; j <= 6; j++) 25 cout << a[i][j] << " "[j == 6]; 26 ok = 1; 27 return; 28 } 29 30 if (num[w[x][y]] == 1) { 31 int v = w[x][y], b = getblock(x, y), p = quyu[v]; 32 if (!rvis[x][p] && !cvis[y][p] && !blockvis[b][p] && !quyuvis[v][p]) { 33 a[x][y] = p; 34 quyu[v] = 0, num[v] = 0; 35 rvis[x][p] = cvis[y][p] = blockvis[b][p] = quyuvis[v][p] = 1; 36 dfs(x, y + 1); 37 rvis[x][p] = cvis[y][p] = blockvis[b][p] = quyuvis[v][p] = 0; 38 quyu[v] = p, num[v] = 1; 39 a[x][y] = 0; 40 } 41 return; 42 } 43 for (int i = 1; i <= min(quyu[w[x][y]] - 1, 6); i++) { 44 int v = w[x][y], b = getblock(x, y), p = i; 45 if (!rvis[x][p] && !cvis[y][p] && !blockvis[b][p] && !quyuvis[v][p]) { 46 a[x][y] = p; 47 quyu[v] -= p, num[v]--; 48 rvis[x][p] = cvis[y][p] = blockvis[b][p] = quyuvis[v][p] = 1; 49 dfs(x, y + 1); 50 rvis[x][p] = cvis[y][p] = blockvis[b][p] = quyuvis[v][p] = 0; 51 quyu[v] += p, num[v]++; 52 a[x][y] = 0; 53 } 54 } 55 } 56 57 int main() { 58 int n; 59 cin >> n; 60 for (int i = 1; i <= n; i++) 61 cin >> quyu[i]; 62 for (int i = 1; i <= 6; i++) 63 for (int j = 1, x; j <= 6; j++) { 64 cin >> x; 65 w[i][j] = x, num[x]++; 66 } 67 dfs(1, 1); 68 }
D：
solver：zyh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 long long sum(long long x) { 4 return x * (x + 1) / 2LL; 5 } 6 int main() { 7 int T; 8 scanf("%d", &T); 9 while (T--) { 10 long long n; 11 scanf("%lld", &n); 12 long long l = -1, r = (long long)sqrt(4 * (n + 1)) + 1; 13 while (l + 1 < r) { 14 long long mid = l + r >> 1; 15 if (sum(mid) <= n) l = mid; 16 else r = mid; 17 } 18 printf("%lld ", l); 19 } 20 return 0; 21 }
E：
solver：lzh
最后绝杀AC，lzhnb
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<int, int> pii; 5 #define ff first 6 #define ss second 7 8 const int N = 1 << 18; 9 const double PI = acos(-1.0); 10 11 struct cpx { 12 double x, y; 13 cpx() {} 14 cpx(double _x, double _y) 15 : x(_x) 16 , y(_y) { 17 } 18 cpx operator+(const cpx &b) const { 19 return cpx(x + b.x, y + b.y); 20 } 21 cpx operator-(const cpx &b) const { 22 return cpx(x - b.x, y - b.y); 23 } 24 cpx operator*(const cpx &b) const { 25 return cpx(x * b.x - y * b.y, x * b.y + y * b.x); 26 } 27 cpx operator/(const int &b) const { 28 return cpx(x / b, y / b); 29 } 30 }; 31 32 int rev[2 * N + 10]; 33 void getrev(int len) { 34 int bit = 0; 35 while ((1 << bit) < len) 36 bit++; 37 for (int i = 0; i < len; i++) 38 rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1)); 39 } 40 41 void fft(cpx x[], int len, int opt) { 42 for (int i = 0; i < len; i++) 43 if (i < rev[i]) 44 swap(x[i], x[rev[i]]); 45 for (int mid = 1; mid < len; mid <<= 1) { 46 cpx tmp(cos(PI / mid), opt * sin(PI / mid)); 47 for (int i = 0, add = mid << 1; i < len; i += add) { 48 cpx base(1, 0); 49 for (int j = i; j < i + mid; j++, base = base * tmp) { 50 cpx a = x[j], b = base * x[j + mid]; 51 x[j] = a + b, x[j + mid] = a - b; 52 } 53 } 54 } 55 56 if (opt == -1) 57 for (int i = 0; i < len; i++) 58 x[i] = x[i] / len; 59 } 60 61 cpx ffta[2 * N + 10], fftb[2 * N + 10]; 62 int solve(int a[], int b[], int lena, int lenb) { 63 int len = 1; 64 while (len < lena + lenb) 65 len <<= 1; 66 getrev(len); 67 68 for (int i = 0; i < lena; i++) 69 ffta[i] = cpx(a[i], 0); 70 for (int i = lena; i < len; i++) 71 ffta[i] = cpx(0, 0); 72 for (int i = 0; i < lenb; i++) 73 fftb[i] = cpx(b[i], 0); 74 for (int i = lenb; i < len; i++) 75 fftb[i] = cpx(0, 0); 76 77 fft(ffta, len, 1), fft(fftb, len, 1); 78 for (int i = 0; i < len; i++) 79 ffta[i] = ffta[i] * fftb[i]; 80 fft(ffta, len, -1); 81 return len; 82 } 83 84 char p[200010], t[200010]; 85 ll dam[N * 2 + 10], ans[N]; 86 const char s1[] = { "RNE" }, s2[] = { "ABC" }; 87 int numa[200010], numb[200010], D[10]; 88 int main() { 89 int a, b, c, h; 90 scanf("%d%d%d%d", &D[0], &D[1], &D[2], &h); 91 scanf("%s%s", p + 1, t + 1); 92 int n = strlen(p + 1), m = strlen(t + 1); 93 for (int i = 0; i < 3; i++) { 94 for (int j = 0; j <= n; j++) 95 numa[j] = numb[j] = 0; 96 for (int j = 1; j <= n; j++) 97 if (p[j] == s1[i]) 98 numa[j]++; 99 for (int j = 1; j <= m; j++) 100 if (t[j] == s2[i]) 101 numb[n - j + 1]++; 102 int len = solve(numa, numb, n + 4, n + 4); 103 for (int j = 0; j < len; j++) 104 dam[j] += (ll)(ffta[j].x + 0.5) * D[i]; 105 //for (int j = 0; j < len; j++) 106 // printf("%d %d %lf ", numa[j], numb[j], ffta[i].x); 107 } 108 int ansn = 0; 109 for (int i = 0; i < 2 * N; i++) { 110 ans[i % n] += dam[i]; 111 } 112 for (int i = 0; i < n; i++) { 113 if (ans[i] >= h) 114 ansn++; 115 //printf("%lld ", ans[i]); 116 } 117 int g = __gcd(ansn, n); 118 printf("%d %d ", ansn / g, n / g); 119 }
F：
solver：czq
很傻的并查集
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 eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 const int MAXN = 1e5 + 10; 15 pair<int, int>edge[MAXN]; 16 int n, m, cnt, fa[MAXN], p[MAXN], ans[MAXN]; 17 18 int main() { 19 scanf("%d%d", &n, &m); cnt = n; 20 for (int i = 1; i <= n; i++) fa[i] = i; 21 for (int i = 1, u, v; i <= m; i++) { 22 scanf("%d%d", &u, &v); 23 edge[i] = mp(u, v); 24 } 25 26 function<int(int)>findFa = [&](int cur)->int{ 27 return fa[cur] == cur ? cur : fa[cur] = findFa(fa[cur]); 28 }; 29 30 for (int i = 1; i <= m; i++) scanf("%d", &p[i]); 31 for (int i = m; i >= 1; i--) { 32 ans[i] = cnt; 33 int fa1 = findFa(edge[p[i]].first), fa2 = findFa(edge[p[i]].second); 34 if (fa1 != fa2) { 35 fa[fa1] = fa2; 36 cnt--; 37 } 38 } 39 for (int i = 1; i <= m; i++) printf("%d ", ans[i]); 40 return 0; 41 }
G：
补：zyh
思路完全正确，队友写挂了，比赛最后三分钟被我对拍拍出错误，但是没时间改
1 #include <bits/stdc++.h> 2 using namespace std; 3 int id[500001]; 4 vector<int> factors[500001]; 5 bool inSet[500001]; 6 long long cnt[1000001]; 7 void getFactors(int x, vector<int> &factor) { 8 int len = sqrt(x); 9 for (int i = 2; i <= len; ++i) 10 if (x % i == 0) { 11 factor.push_back(i); 12 if (i * i != x) factor.push_back(x / i); 13 } 14 factor.push_back(x); 15 } 16 void updateCnt(vector<int> &factor, int d) { 17 for (auto x : factor) { 18 cnt[x] += d; 19 } 20 } 21 void getPrime(int n, vector<int> &prime) { 22 prime.clear(); 23 int len = sqrt(n); 24 for (int i = 2; i <= len; ++i) 25 if (n % i == 0) { 26 prime.push_back(i); 27 while (n % i == 0) n /= i; 28 } 29 if (n > 1) prime.push_back(n); 30 } 31 void dfs(int st, int sel, int product, long long &s, vector<int> &prime, int len) { 32 if (sel > 0) { 33 if (sel & 1) s += cnt[product]; 34 else s -= cnt[product]; 35 } 36 for (int i = st; i < len; ++i) { 37 dfs(i + 1, sel + 1, product * prime[i], s, prime, len); 38 } 39 } 40 long long howMany(int x) { 41 vector<int> prime; 42 getPrime(x, prime); 43 int len = prime.size(); 44 long long rnt = 0; 45 dfs(0, 0, 1, rnt, prime, len); 46 //cout<<"howMany: "<<rnt<<endl; 47 return rnt; 48 } 49 int main() { 50 int n, T; 51 scanf("%d%d", &n, &T); 52 for (int i = 0; i < n; ++i) { 53 scanf("%d", &id[i]); 54 getFactors(id[i], factors[i]); 55 //cout<<"factor: "; 56 //for (auto x:factors[i]) cout<<x<<' '; 57 //cout<<endl; 58 } 59 long long ans = 0; 60 long long sizeOfSet = 0; 61 while (T--) { 62 int t; 63 scanf("%d", &t); 64 --t; 65 if (inSet[t]) { 66 updateCnt(factors[t], -1); 67 sizeOfSet--; 68 ans -= sizeOfSet - howMany(id[t]); 69 } else { 70 ans += sizeOfSet - howMany(id[t]); 71 updateCnt(factors[t], 1); 72 sizeOfSet++; 73 } 74 inSet[t] = !inSet[t]; 75 printf("%lld ", ans); 76 } 77 }
H：
solver：czq
很傻的概率题
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 eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 double n, a, p; 15 16 int main() { 17 scanf("%lf%lf%lf", &n, &a, &p); 18 printf("%.8f ", (a * p - a * (100.0 - p)) / 100.0 + n); 19 return 0; 20 }
I：
solver：lzh、zyh
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<int, int> pii; 5 #define ff first 6 #define ss second 7 const int mod = 1e9 + 7; 8 9 ll mul(pii a, pii b, pii c) { 10 c.ff = c.ff - a.ff, c.ss = c.ss - a.ss; 11 a.ff = b.ff - a.ff, a.ss = b.ss - a.ss; 12 return 1ll * a.ff * c.ss - 1ll * a.ss * c.ff; 13 } 14 15 pii a[1010], b[1010], c[1010]; 16 int main() { 17 int s, r, m, n; 18 cin >> s >> r >> m; 19 for (int i = 1; i <= s; i++) 20 cin >> a[i].ff >> a[i].ss; 21 for (int i = 1; i <= r; i++) 22 cin >> b[i].ff >> b[i].ss; 23 for (int i = 1; i <= m; i++) 24 cin >> c[i].ff >> c[i].ss; 25 a[s + 1] = a[1], b[r + 1] = b[1], c[m + 1] = c[1]; 26 cin >> n; 27 while (n--) { 28 pii x; 29 cin >> x.ff >> x.ss; 30 int gg = 0; 31 for (int i = 1; i <= s; i++) 32 if (mul(a[i], a[i + 1], x) < 0) { 33 gg++; 34 break; 35 } 36 if (!gg) { 37 cout << "Sheena "; 38 continue; 39 } 40 gg = 0; 41 for (int i = 1; i <= r; i++) 42 if (mul(b[i], b[i + 1], x) < 0) { 43 gg++; 44 break; 45 } 46 if (!gg) { 47 cout << "Rose "; 48 continue; 49 } 50 gg = 0; 51 for (int i = 1; i <= m; i++) 52 if (mul(c[i], c[i + 1], x) < 0) { 53 gg++; 54 break; 55 } 56 if (!gg) { 57 cout << "Maria "; 58 continue; 59 } 60 cout << "Outside "; 61 } 62 }
J：
solver：czq
树状数组
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 eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 class FenwickTree { 15 private: 16 // 存储的信息类型 17 typedef int StoreType; 18 int n; 19 // num用于存储元素值，用于实现单点修改、区间查询 20 // diff用于存储差分，用于实现区间修改 21 vector<StoreType>num, diff; 22 public: 23 FenwickTree(): n(0) {} 24 FenwickTree(int _n) { 25 n = _n; 26 num.assign(n, 0); 27 } 28 // 单点修改 29 void add(int pos, int val) { 30 for (; pos < n; pos |= pos + 1) num[pos] += val; 31 } 32 // 查询区间[0..pos]的信息 33 StoreType sum(int pos) { 34 StoreType ret = 0; 35 for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ret += num[pos]; 36 return ret; 37 } 38 // 查询区间[l..r]的和 39 StoreType querySum(int l, int r) { 40 return sum(r) - sum(l - 1); 41 } 42 // 区间修改，给区间[l..r]里的元素都加上val 43 void intervalFix(int l, int r, int val) { 44 add(l, val); add(r + 1, -val); 45 } 46 }; 47 48 const int MAXN = 1e5 + 10; 49 int n, a[MAXN]; 50 FenwickTree fwt(MAXN * 10); 51 52 int main() { 53 scanf("%d", &n); 54 for (int i = 1; i <= n; i++) { 55 scanf("%d", &a[i]); 56 a[i]++; 57 printf("%d ", n - (i - fwt.sum(a[i] - 1)) + 1); 58 fwt.add(a[i], 1); 59 } 60 return 0; 61 }
K：
solver：lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<int, int> pii; 5 #define ff first 6 #define ss second 7 8 int dis[100010][3]; 9 struct edge { 10 int u, v, w; 11 edge(int _u = 0, int _v = 0, int _w = 0) { 12 u = _u, v = _v, w = _w; 13 } 14 } e[200010]; 15 int head[100010], num = 0, vis[200010]; 16 void addedge() { 17 int u, v, w; 18 cin >> u >> v >> w; 19 e[num] = edge(head[u], v, w); 20 head[u] = num++; 21 e[num] = edge(head[v], u, w); 22 head[v] = num++; 23 } 24 struct node { 25 int val, v, add; 26 node(int _val = 0, int _v = 0, int _add = 0) { 27 val = _val, v = _v; 28 add = _add; 29 } 30 bool operator<(const node &b) const { 31 return val > b.val; 32 } 33 }; 34 int main() { 35 ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); 36 memset(dis, 0x3f, sizeof dis); 37 memset(head, -1, sizeof head); 38 39 int n, m; 40 cin >> n >> m; 41 for (int i = 1; i <= m; i++) { 42 addedge(); 43 } 44 priority_queue<node> q; 45 q.push(node{ 0, 1, 0 }); 46 dis[1][0] = 0; 47 while (!q.empty()) { 48 node x = q.top(); 49 q.pop(); 50 for (int i = head[x.v]; i != -1; i = e[i].u) { 51 if ((vis[i] & x.add) == 0 && (vis[i ^ 1] & x.add) == 0) { 52 vis[i] |= x.add, vis[i ^ 1] |= x.add; 53 for (int j = 0; j < 3; j++) { 54 int tmp = dis[e[i].v][j]; 55 tmp = min(tmp, dis[x.v][(j + 2) % 3] + e[i].w); 56 if (e[i].v != n) { 57 tmp = min(tmp, dis[x.v][j] + e[i].w * 3); 58 tmp = min(tmp, dis[x.v][(j + 1) % 3] + e[i].w * 5); 59 } 60 if (e[i].v != n && dis[e[i].v][j] != tmp) { 61 if ((vis[i] & 1) == 0) 62 q.push(node{ dis[e[i].v][0], e[i].v, 1 }); 63 if ((vis[i] & 2) == 0) 64 q.push(node{ dis[e[i].v][1], e[i].v, 2 }); 65 if ((vis[i] & 4) == 0) 66 q.push(node{ dis[e[i].v][2], e[i].v, 4 }); 67 } 68 dis[e[i].v][j] = tmp; 69 } 70 } 71 } 72 } 73 pair<int, string> ans[10]; 74 ans[1] = { dis[n][0], "me" }; 75 ans[2] = { dis[n][1], "Gon" }; 76 ans[3] = { dis[n][2], "Killua" }; 77 sort(ans + 1, ans + 1 + 3); 78 for (int i = 1; i <= 3; i++) 79 cout << ans[i].ss << endl; 80 }
L：
solver：zyh、lzh、czq
读错题把队友坑了，锅++
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> point; 4 vector<point> pts; 5 bool check(point a, point b) { 6 return (a.first < b.first && abs(a.first - b.first) + abs(a.second - b.second) <= 5); 7 } 8 vector<int> edges[200001]; 9 int dp[200001]; 10 bool vis[200001]; 11 bool reach0[200001]; 12 map<int, map<int, int>> Map; 13 void dfs(int u) { 14 vis[u] = true; 15 dp[u] = 1; 16 for (auto v : edges[u]) { 17 if (!vis[v]) dfs(v); 18 if (reach0[v]) dp[u] = max(dp[u], dp[v] + 1); 19 reach0[u] |= reach0[v]; 20 } 21 } 22 int main() { 23 int n, l; 24 scanf("%d%d", &n, &l); 25 pts.push_back(point(0, 0)); 26 Map[0][0] = 0; 27 for (int i = 0; i < n; ++i) { 28 int x, y; 29 scanf("%d%d", &x, &y); 30 pts.push_back(point(x, y)); 31 Map[x][y] = i + 1; 32 } 33 for (int i = 0; i <= n; ++i) { 34 int x = pts[i].first; 35 int y = pts[i].second; 36 for (int j = 1; j <= 5; ++j) { 37 if (Map.find(x + j) == Map.end()) continue; 38 auto iter = Map[x + j].lower_bound(y - (5 - j)); 39 while (iter != Map[x + j].end() && iter->first <= y + 5 - j) { 40 edges[iter->second].push_back(i); 41 iter++; 42 } 43 } 44 } 45 reach0[0] = true; 46 for (int i = 0; i <= n; ++i) 47 if (!vis[i]) dfs(i); 48 point end = {l, 0}; 49 int ans = 0; 50 for (int i = 0; i <= n; ++i) { 51 if (reach0[i] && check(pts[i], end)) ans = max(ans, dp[i] - 1); 52 } 53 printf("%d", ans); 54 }