标签：variables etc names eject 12c push res bool isp
题目很简单。差点10题，非常可惜。
A：
签到题+4不应该。
solver：lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 typedef long long ll; 5 #define ff first 6 #define ss second 7 8 const int MAXSTRLEN = 500010; 9 int maxlen0 = 0, maxlen1 = 0; 10 char ch[2 * MAXSTRLEN]; 11 int p[2 * MAXSTRLEN]; 12 void manacher(char str[], int len) { 13 for (int i = 0; i < len; i++) { 14 ch[2 * i + 1] = ‘#‘; 15 ch[2 * i + 2] = str[i]; 16 } 17 ch[0] = ‘$‘; 18 ch[2 * len + 1] = ‘#‘; 19 ch[2 * len + 2] = ‘\0‘; 20 len = len * 2 + 2; 21 p[0] = p[1] = 1; 22 int id = 1, mx = 2; 23 for (int i = 2; i < len; i++) { 24 int j = min(p[2 * id - i], mx - i); 25 while (ch[i - j] == ch[i + j]) 26 j++; 27 p[i] = j; 28 if (i + p[i] > mx) { 29 id = i; 30 mx = i + p[i]; 31 } 32 if ((p[i] - 1) % 2 == 0) 33 maxlen0 = max(maxlen0, p[i] - 1); 34 else 35 maxlen1 = max(maxlen1, p[i] - 1); 36 } 37 } 38 char s[MAXSTRLEN]; 39 int main() { 40 int n, m; 41 scanf("%d%d%s", &n, &m, s); 42 manacher(s, n); 43 if ((m % 2 == 0 && maxlen0 >= m) || (m % 2 == 1 && maxlen1 >= m)) { 44 printf("Accept\n"); 45 } else 46 printf("Reject\n"); 47 }
B：
solver：zyh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 struct Segment_Tree { 4 //constants and innerclass 5 struct node { 6 int l, r, lc, rc; 7 long long val; 8 }; 9 static const int N = 500000; 10 static const int root = 0; 11 12 //variables 13 int num; 14 node tr[2 * N]; 15 16 //methods 17 void setRange(int l, int r, int now = root) { 18 num = 0; 19 build(l, r, now); 20 } 21 void build(int l, int r, int now) { 22 tr[now].l = l; tr[now].r = r; tr[now].val = 0; 23 if (l < r) { 24 int mid = (l + r) >> 1; 25 ++num; tr[now].lc = num; 26 ++num; tr[now].rc = num; 27 build(l, mid, tr[now].lc); 28 build(mid + 1, r, tr[now].rc); 29 } 30 } 31 32 void pushup(node &f, node &lc, node &rc) { 33 f.val = lc.val + rc.val; 34 } 35 36 //warning: don‘t invoke both update and updateR in single segment tree, it may cause error 37 void update(int pos, long long val, int now = root) { 38 int mid = (tr[now].l + tr[now].r) >> 1; 39 if (tr[now].l == tr[now].r) { 40 tr[now].val = val; 41 return; 42 } else if (pos <= mid) update(pos, val, tr[now].lc); 43 else update(pos, val, tr[now].rc); 44 //write parent update here 45 pushup(tr[now], tr[tr[now].lc], tr[tr[now].rc]); 46 } 47 48 long long query(int l, int r, int now = root) { 49 if (tr[now].l == l && tr[now].r == r) return tr[now].val; 50 else { 51 //if (tr[now].cover!=0) pushdown(tr[now],tr[tr[now].lc],tr[tr[now].rc]); 52 int mid = (tr[now].l + tr[now].r) >> 1; 53 if (r <= mid) return query(l, r, tr[now].lc); 54 else if (l > mid) return query(l, r, tr[now].rc); 55 else return query(l, mid, tr[now].lc) + query(mid + 1, r, tr[now].rc); 56 } 57 } 58 59 }; 60 Segment_Tree tr; 61 map<string, long long> Map; 62 string str[1000001]; 63 char ch[10001]; 64 int main() { 65 int n, m, q; 66 scanf("%d%d%d", &n, &m, &q); 67 for (int i = 1; i <= n; ++i) { 68 scanf("%s", ch); 69 str[i] = ch; 70 } 71 for (int i = 0; i < m; ++i) { 72 long long v; 73 scanf("%s%lld", ch, &v); 74 Map[string(ch)] = v; 75 } 76 tr.setRange(1, n); 77 for (int i = 1; i <= n; ++i) { 78 tr.update(i, Map[str[i]]); 79 } 80 while (q--) { 81 int op; 82 scanf("%d", &op); 83 if (op == 1) { 84 int x; 85 scanf("%d%s", &x, ch); 86 tr.update(x, Map[string(ch)]); 87 } else { 88 int l, r; 89 scanf("%d%d", &l, &r); 90 long long s = tr.query(l, r); 91 if (s <= 30 * (r - l + 1)) printf("NO\n"); 92 else printf("YES\n"); 93 } 94 } 95 }
D：
solver：lzh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<double, int> pii; 5 #define ff first 6 #define ss second 7 #define mp make_pair 8 9 pii p[100010]; 10 int check(double x, int n) { 11 int cur = 1; 12 while (cur <= n) { 13 ll add = 0; 14 while (p[cur].ff - x <= 0) 15 add += p[cur++].ss; 16 if (add == 0) 17 return 0; 18 x += add; 19 } 20 return 1; 21 } 22 int main() { 23 int n, x, y; 24 scanf("%d%d%d", &n, &x, &y); 25 for (int i = 1; i <= n; i++) { 26 int u, v, w; 27 scanf("%d%d%d", &u, &v, &w); 28 p[i] = mp(sqrt(1ll * (u - x) * (u - x) + 1ll * (v - y) * (v - y)) - w, w); 29 } 30 sort(p + 1, p + 1 + n); 31 32 double l = max(0.0, p[1].ff), r = p[n].ff, eps = 1e-8; 33 double ans = r; 34 while (r - l > eps) { 35 double mid = (l + r) / 2.0; 36 if (check(mid, n)) { 37 r = mid; 38 ans = min(ans, mid); 39 } else 40 l = mid; 41 } 42 printf("%.10lf\n", ans); 43 }
E：
solver：zyh
1 #include <bits/stdc++.h> 2 using namespace std; 3 int li[] = {6, 28, 496, 8128}; 4 int len = 4; 5 inline int read() { 6 int rnt = 0; 7 int sign = 0; 8 char ch = 0; 9 while (!isdigit(ch)) { 10 sign |= ch == ‘-‘; 11 ch = getchar(); 12 } 13 while (isdigit(ch)) rnt = (rnt << 3) + (rnt << 1) + (ch ^ 48), ch = getchar(); 14 return sign ? -rnt : rnt; 15 } 16 inline void print(int x) { 17 if (x < 0) { 18 putchar(‘-‘); 19 x = -x; 20 } 21 if (x > 9) print(x / 10); 22 putchar(x % 10 + ‘0‘); 23 } 24 int main() { 25 int n = read(); 26 while (n--) { 27 int k = read(); 28 int ans; 29 for (ans = len - 1; ans >= 0; --ans) 30 if (li[ans] <= k) break; 31 if (ans < 0) print(-1); 32 else print(li[ans]); 33 putchar(‘\n‘); 34 } 35 }
F：
solver：lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 typedef long long ll; 5 #define ff first 6 #define ss second 7 8 struct node { 9 int l, r, h; 10 int add; 11 node() {} 12 node(int _l, int _r, int _h, int _add): l(_l), r(_r), h(_h), add(_add) {} 13 }; 14 15 const int N = 2e5 + 10; 16 int x[N]; 17 ll sum[N << 2]; 18 node q[N]; 19 int mark[N << 2]; 20 21 void pushup(int n, int left, int right) { 22 if (mark[n])sum[n] = x[right + 1] - x[left]; 23 else if (left == right)sum[n] = 0; 24 else sum[n] = sum[n << 1] + sum[n << 1 | 1]; 25 } 26 27 void pushdown(int n) { 28 mark[n << 1] += mark[n], mark[n << 1 | 1] += mark[n]; 29 mark[n] = 0; 30 } 31 32 void update(int l, int r, int add, int n, int left, int right) { 33 if (l <= left && right <= r) { 34 mark[n] += add; 35 pushup(n, left, right); 36 return; 37 } 38 int mid = left + right >> 1; 39 if (l <= mid)update(l, r, add, n << 1, left, mid); 40 if (mid < r)update(l, r, add, n << 1 | 1, mid + 1, right); 41 pushup(n, left, right); 42 } 43 44 int main() { 45 int n; 46 scanf("%d", &n); 47 for (int i = 1; i <= n; i++) { 48 int l, r, h; scanf("%d%d%d", &l, &r, &h); 49 x[2 * i - 1] = l; x[2 * i] = r; 50 q[2 * i - 1] = node(l, r, 0, 1); 51 q[2 * i] = node(l, r, h, -1); 52 } 53 sort(x + 1, x + 1 + 2 * n); 54 sort(q + 1, q + 1 + 2 * n, [](node a, node b) { 55 return a.h < b.h; 56 }); 57 int cnt = 1; 58 for (int i = 2; i <= 2 * n; i++) 59 if (x[i - 1] != x[i])x[++cnt] = x[i]; 60 ll ans = 0; 61 q[2 * n + 1].h = q[2 * n].h; 62 63 for (int i = 1; i <= 2 * n; i++) { 64 int l = lower_bound(x + 1, x + 1 + cnt, q[i].l) - x; 65 int r = lower_bound(x + 1, x + 1 + cnt, q[i].r) - x - 1; 66 update(l, r, q[i].add, 1, 1, cnt); 67 ans += sum[1] * (q[i + 1].h - q[i].h); 68 } 69 printf("%lld\n", ans); 70 }
G：
solver：lzh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 typedef long long ll; 5 #define ff first 6 #define ss second 7 8 map<string, int>name, prob; 9 char strname[50010][25]; 10 int pts[50010]; 11 set<int>submit[50010]; 12 int main() { 13 int c, p, s; scanf("%d%d%d", &c, &p, &s); 14 int cname = 0, cprob = 0; 15 char str[50]; 16 for (int i = 1; i <= c; i++) { 17 scanf("%s", strname[i]); 18 name[strname[i]] = ++cname; 19 } 20 for (int i = 1; i <= p; i++) { 21 int x; 22 scanf("%s%d", str, &x); 23 prob[str] = ++cprob; pts[cprob] = x; 24 } 25 for (int i = 1; i <= s; i++) { 26 int numname = -1, numprob = -1; 27 scanf("%s", str); 28 numname = name[str]; 29 scanf("%s", str); 30 numprob = prob[str]; 31 scanf("%s", str); 32 if (strlen(str) == 2 && str[0] == ‘A‘ && str[1] == ‘C‘) { 33 if (numname && numprob) { 34 submit[numname].insert(numprob); 35 } 36 } 37 } 38 for (int i = 1; i <= c; i++) { 39 int ans = 0; 40 for (auto j : submit[i])ans += pts[j]; 41 printf("%s %d\n", strname[i], ans); 42 } 43 }
H：
solver：lzh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef pair<ll, int> pii; 5 #define ff first 6 #define ss second 7 #define mp make_pair 8 9 const int N = 1e5 + 10; 10 vector<pii> v[N]; 11 int T[N], vis[N]; 12 ll ans[N]; 13 int main() { 14 int n, m, k; 15 scanf("%d%d%d", &n, &m, &k); 16 for (int i = 1; i <= m; i++) { 17 int x, y, z; 18 scanf("%d%d%d", &x, &y, &z); 19 v[x].push_back(mp(y, z)); 20 v[y].push_back(mp(x, z)); 21 } 22 for (int i = 1; i <= n; i++) 23 scanf("%d", &T[i]); 24 25 priority_queue<pii, vector<pii>, greater<pii>> pq; 26 for (int i = 1; i <= k; i++) { 27 int x; 28 scanf("%d", &x); 29 pq.push(mp(T[x], x)); 30 } 31 32 while (!pq.empty()) { 33 pii x = pq.top(); 34 pq.pop(); 35 if (vis[x.ss]) 36 continue; 37 vis[x.ss]++; 38 ans[x.ss] = x.ff; 39 40 for (auto i : v[x.ss]) 41 if (!vis[i.ff]) 42 pq.push(mp(x.ff + i.ss + T[i.ff], i.ff)); 43 } 44 for (int i = 1; i <= n; i++) 45 printf("%lld\n", ans[i]); 46 }
I：
solver：zyh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 struct point { 4 int x, v; 5 point() {} 6 point(int _x, int _v) { 7 x = _x; 8 v = _v; 9 } 10 bool operator<(const point &b)const { 11 return v < b.v; 12 } 13 }; 14 point pts[2000000]; 15 int f[2000000]; 16 int Size[2000000]; 17 int a[2000000]; 18 int find(int u) { 19 if (f[u] != u) f[u] = find(f[u]); 20 return f[u]; 21 } 22 int Union(int u, int v) { 23 u = find(u); 24 v = find(v); 25 if (u != v) { 26 Size[u] += Size[v]; 27 f[v] = u; 28 } 29 return u; 30 } 31 int main() { 32 int s, n, m; 33 scanf("%d%d%d", &s, &n, &m); 34 int len = n * m; 35 for (int i = 0; i < len; ++i) { 36 scanf("%d", &a[i]); 37 pts[i] = point(i, a[i]); 38 } 39 sort(pts, pts + len); 40 for (int i = 0; i < len; ++i) { 41 f[i] = i; Size[i] = 1; 42 } 43 int maxsize = 0; 44 for (int k = len - 1; k >= 0; --k) { 45 int i = pts[k].x; 46 int v = pts[k].v; 47 if (i - 1 >= 0 && (i / m) == ((i - 1) / m) && a[i - 1] >= v) Union(i, i - 1); 48 if (i + 1 < len && (i / m) == ((i + 1) / m) && a[i + 1] >= v) Union(i, i + 1); 49 if (i + m < len && a[i + m] >= v) Union(i, i + m); 50 if (i - m >= 0 && a[i - m] >= v) Union(i, i - m); 51 maxsize = max(maxsize, Size[find(i)]); 52 if (maxsize >= s) { 53 printf("%d\n", v); 54 return 0; 55 } 56 } 57 }
J：
solver：lzh
差一点就做出来了，非常难顶
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 typedef long long ll; 5 #define ff first 6 #define ss second 7 #define mp make_pair 8 9 vector<pii> v[100010]; 10 ll dp[100010][10]; 11 const ll mod = 1e9 + 7; 12 int baoli2[20][2] = { 1, 2, 1, 3, 1, 4, 1, 5, 2, 1, 2, 3, 2, 4, 2, 5, 3, 1, 3, 2, 3, 4, 3, 5, 4, 1, 4, 2, 4, 3, 4, 5, 5, 1, 5, 2, 5, 3, 5, 4 }; 13 int baoli3[60][3] = { 1, 2, 3, 1, 2, 4, 1, 2, 5, 1, 3, 2, 1, 3, 4, 1, 3, 5, 1, 4, 2, 1, 4, 3, 1, 4, 5, 1, 5, 2, 1, 5, 3, 1, 5, 4, 2, 1, 3, 2, 1, 4, 2, 1, 5, 2, 3, 1, 2, 3, 4, 2, 3, 5, 2, 4, 1, 2, 4, 3, 2, 4, 5, 2, 5, 1, 2, 5, 3, 2, 5, 4, 3, 1, 2, 3, 1, 4, 3, 1, 5, 3, 2, 1, 3, 2, 4, 3, 2, 5, 3, 4, 1, 3, 4, 2, 3, 4, 5, 3, 5, 1, 3, 5, 2, 3, 5, 4, 4, 1, 2, 4, 1, 3, 4, 1, 5, 4, 2, 1, 4, 2, 3, 4, 2, 5, 4, 3, 1, 4, 3, 2, 4, 3, 5, 4, 5, 1, 4, 5, 2, 4, 5, 3, 5, 1, 2, 5, 1, 3, 5, 1, 4, 5, 2, 1, 5, 2, 3, 5, 2, 4, 5, 3, 1, 5, 3, 2, 5, 3, 4, 5, 4, 1, 5, 4, 2, 5, 4, 3 }; 14 15 void dfs(int x, int pre) { 16 int col = 0; 17 vector<int> nex; 18 for (auto i : v[x]) 19 if (i.ff != pre) { 20 dfs(i.ff, x); 21 nex.push_back(i.ff); 22 } else 23 col = i.ss; 24 int sz = v[x].size(); 25 if (pre == -1) 26 return; 27 28 if (sz == 1) { 29 if (v[x][0].ss) 30 dp[x][v[x][0].ss] = 1; 31 else 32 for (int i = 1; i <= 5; i++) 33 dp[x][i] = 1; 34 return; 35 } else if (sz == 2) { 36 for (int i = 0; i < 20; i++) 37 if (!col || col == baoli2[i][1]) { 38 dp[x][baoli2[i][1]] = (dp[x][baoli2[i][1]] + dp[nex[0]][baoli2[i][0]]) % mod; 39 } 40 } else if (sz == 3) { 41 for (int i = 0; i < 60; i++) 42 if (!col || col == baoli3[i][2]) { 43 dp[x][baoli3[i][2]] = (dp[x][baoli3[i][2]] + dp[nex[0]][baoli3[i][0]] * dp[nex[1]][baoli3[i][1]] % mod) % mod; 44 } 45 } else { 46 int a[10]; 47 for (int i = 1; i <= 5; i++) 48 a[i] = i; 49 do { 50 if (!col || col == a[sz]) { 51 ll tmp = 1; 52 for (int i = 1; i <= sz - 1; i++) 53 tmp = tmp * dp[nex[i - 1]][a[i]] % mod; 54 55 dp[x][a[sz]] = (dp[x][a[sz]] + tmp) % mod; 56 } 57 } while (next_permutation(a + 1, a + 1 + 5)); 58 } 59 } 60 int main() { 61 int n; 62 scanf("%d", &n); 63 for (int i = 1; i <= n - 1; i++) { 64 int a, b, c; 65 scanf("%d%d%d", &a, &b, &c); 66 v[a].push_back(mp(b, c)); 67 v[b].push_back(mp(a, c)); 68 } 69 for (int i = 1; i <= n; i++) 70 if (v[i].size() > 5) { 71 printf("0\n"); 72 return 0; 73 } 74 if (n == 1) { 75 printf("1\n"); 76 return 0; 77 } 78 if (n == 2) { 79 if (v[1][0].ss) 80 printf("1\n"); 81 else 82 printf("5\n"); 83 return 0; 84 } 85 dfs(1, -1); 86 87 ll ans = 0; 88 vector<int> nex; 89 for (auto i : v[1]) 90 nex.push_back(i.ff); 91 if (v[1].size() == 1) { 92 for (int i = 1; i <= 5; i++) { 93 ans = (ans + dp[nex[0]][i]) % mod; 94 } 95 } else if (v[1].size() == 2) { 96 for (int i = 0; i < 20; i++) { 97 ll tmp = 1; 98 for (int j = 0; j < 2; j++) 99 tmp = tmp * dp[nex[j]][baoli2[i][j]] % mod; 100 ans = (ans + tmp) % mod; 101 } 102 } else if (v[1].size() == 3) { 103 for (int i = 0; i < 60; i++) { 104 ll tmp = 1; 105 for (int j = 0; j < 3; j++) 106 tmp = tmp * dp[nex[j]][baoli3[i][j]] % mod; 107 ans = (ans + tmp) % mod; 108 } 109 } else { 110 int a[10], sz = v[1].size(); 111 for (int i = 1; i <= 5; i++) 112 a[i] = i; 113 do { 114 ll tmp = 1; 115 for (int i = 1; i <= sz; i++) 116 tmp = tmp * dp[nex[i - 1]][a[i]] % mod; 117 ans = (ans + tmp) % mod; 118 } while (next_permutation(a + 1, a + 1 + 5)); 119 } 120 121 printf("%lld\n", ans); 122 }
K：
solver：lzh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 typedef long long ll; 5 #define ff first 6 #define ss second 7 8 const int N = 1e6 + 10; 9 vector<ll> dp[N]; 10 vector<int> a[N]; 11 int main() { 12 int n, h; scanf("%d%d", &n, &h); 13 for (int i = 1; i <= n; i++) { 14 a[i].resize(h + 1); 15 dp[i].resize(h + 1); 16 for (int j = 1; j <= h; j++) 17 scanf("%d", &a[i][j]); 18 dp[i][1] = a[i][1]; 19 } 20 21 a[0].resize(h + 1); a[n + 1].resize(h + 1); 22 dp[0].resize(h + 1); dp[n + 1].resize(h + 1); 23 for (int i = 1; i <= h; i++)a[0][i] = a[n + 1][i] = dp[0][i] = dp[n + 1][i] = 0; 24 for (int j = 2; j <= h; j++) { 25 for (int i = 1; i <= n; i++) { 26 dp[i][j] = a[i][j] + max(max(dp[i - 1][j - 1], dp[i + 1][j - 1]), dp[i][j - 1]); 27 } 28 } 29 ll maxx = 0; 30 for (int i = 1; i <= n; i++)maxx = max(maxx, dp[i][h]); 31 printf("%lld\n", maxx); 32 }
UFPE Starters Final Try-Outs 2020
标签：variables etc names eject 12c push res bool isp
原文地址：https://www.cnblogs.com/JHSeng/p/12193790.html
