Codeforces Round #561 (div. 2)
时间太晚就没有打，随便看了一下题，当了嘴巴选手。结果发现题目简单到超出想象。
题目链接：https://codeforces.com/contest/1166
A：
上来就卡题意，其实用map维护一下直接秒杀。
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define pb push_back 4 #define mp make_pair 5 #define sot(a,b) sort(a+1,a+1+b) 6 #define rep(i,a,b) for (int i=a;i<=b;i++) 7 #define eps 1e-8 8 #define int_inf (1<<30)-1 9 #define ll_inf (1LL<<62)-1 10 #define lson curPos<<1 11 #define rson curPos<<1|1 12 13 using namespace std; 14 15 map<char, int>m; 16 int n; 17 string s; 18 19 int main() 20 { 21 m.clear(); 22 cin >> n; 23 while (n--) 24 { 25 cin >> s; 26 m[s[0]]++; 27 } 28 int ans = 0; 29 for (auto i : m) 30 { 31 if (i.second <= 2) continue; 32 int tmp = i.second / 2; 33 ans += (tmp - 1) * tmp / 2; 34 if (i.second & 1) tmp++; 35 ans += (tmp - 1) * tmp / 2; 36 } 37 cout << ans << endl; 38 return 0; 39 }
B：
只要能把k分解为两个大于等于5的数相乘，暴力填上aeiou就完事了。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson curPos<<1 15 #define rson curPos<<1|1 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 int k, n, m, flag = 0; 21 string s = "aeiou"; 22 23 int main() 24 { 25 scanf("%d", &k); 26 rep1(i, 1, k) 27 if (k % i == 0 && i >= 5 && k / i >= 5) 28 { 29 n = i, m = k / i, flag = 1; break; 30 } 31 if (!flag) return puts("-1"), 0; 32 rep0(i, 0, n) rep0(j, 0, m) printf("%c", s[(i + j) % 5]); 33 puts(""); 34 return 0; 35 }
C：
给定一维坐标系上的n个数(点)，问有多少对数x,y满足线段 [abs(x-y),abs(x+y)]能覆盖线段[abs(x),abs(y)]。
看似麻烦。但是留意到被覆盖的线段是[abs(x),abs(y)]而不是[x,y]，所以这道题跟正负没有任何关系，直接取绝对值即可。判断大的线段能不能覆盖小的线段，就要看max(x,y)/min(x,y)是否大于2，如果大于2则不能覆盖。显然二分查找。
于是时间复杂度O(nlogn)，是这段时间cf contest出过最傻逼的C题了。(赛后看了一眼队友代码，怎么你们还分类大讨论啊？？？)
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson curPos<<1 15 #define rson curPos<<1|1 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 2e5 + 10; 21 int a[maxn], n; 22 ll ans = 0; 23 24 int main() 25 { 26 scanf("%d", &n); 27 rep1(i, 1, n) scanf("%d", &a[i]), a[i] = abs(a[i]); 28 sot(a, n); 29 rep0(i, 1, n) ans += upper_bound(a + i, a + 1 + n, 2 * a[i]) - a - i - 1; 30 printf("%lld\n", ans); 31 return 0; 32 }
D：
直接枚举序列长度+二分。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson curPos<<1 15 #define rson curPos<<1|1 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 1e2 + 10; 21 ll n, m, q, a, b, tot = 1, pre[maxn], used[maxn]; 22 23 int main() 24 { 25 scanf("%d", &q); 26 pre[1] = 1; 27 rep1(i, 2, 50) //弄个前缀和 28 { 29 pre[i] = tot; 30 tot += pre[i]; 31 } 32 while (q--) 33 { 34 scanf("%lld%lld%lld", &a, &b, &m); 35 bool flag = false; 36 rep1(k, 1, 50) //枚举序列长度 37 { 38 for (int j = k - 1; j > 0; j--) used[j] = 1; //init 39 ll curr = pre[k] * a; 40 rep0(j, 1, k) curr += pre[j]; 41 if (curr > b) break; //直接计算最后一项，若不符合，之后都无解 42 for (int j = k - 1; j > 0; j--) 43 { 44 ll l = 0, r = m - 1; //binary search 45 while (l <= r) 46 { 47 ll mid = l + r >> 1; 48 if (curr + mid * pre[j] > b) r = mid - 1; 49 else l = mid + 1; 50 } 51 used[j] = --l + 1; 52 curr += pre[j] * l; 53 } 54 if (curr == b) //有解 55 { 56 flag = true; 57 ll curr = a, currsum = 0; 58 printf("%d", k); 59 rep1(j, 1, k) 60 { 61 printf(" %lld", curr); 62 currsum += curr; 63 curr = currsum + used[k - j]; 64 } 65 puts(""); 66 break; 67 } 68 } 69 if (!flag) puts("-1"); 70 } 71 return 0; 72 }
E：
数论题，完全莫得想法，据说是个结论题(反正数论全靠队友2333)。贴队友代码(via. rsqppp)好了。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 5 using namespace std; 6 int f[60][15000]; 7 int main() 8 { 9 int m,n; 10 scanf("%d%d",&m,&n); 11 for(int i = 1;i <= m;i++){ 12 int num; 13 scanf("%d",&num); 14 for(int j = 1;j <= num;j++){ 15 int x; 16 scanf("%d",&x); 17 f[i][x] = 1; 18 } 19 } 20 for(int i = 1;i <= m;i++){ 21 for(int j = 1;j <= m;j++){ 22 int flag = 0; 23 for(int k = 1;k <= n;k++){ 24 if(f[i][k] && f[j][k]){ 25 flag = 1; 26 break; 27 } 28 } 29 if(flag == 0){ 30 printf("impossible\n"); 31 return 0; 32 } 33 } 34 } 35 printf("possible\n"); 36 return 0; 37 }
F：
其实不是很想做，不过看到是数据结构题，还是要主动背起这口锅……就是个并查集维护动态图的题。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define sot(a,b) sort(a+1,a+1+b) 8 #define rep1(i,a,b) for(int i=a;i<=b;++i) 9 #define rep0(i,a,b) for(int i=a;i<b;++i) 10 #define eps 1e-8 11 #define int_inf 0x3f3f3f3f 12 #define ll_inf 0x7f7f7f7f7f7f7f7f 13 #define lson curPos<<1 14 #define rson curPos<<1|1 15 /* namespace */ 16 using namespace std; 17 /* header end */ 18 19 const int maxn = 1e5 + 10; 20 int n, m, c, q, fa[maxn]; 21 map<int, int>mp[maxn]; 22 set<int>own[maxn]; 23 24 int findFa(int v) 25 { 26 return fa[v] < 0 ? v : fa[v] = findFa(fa[v]); 27 } 28 29 void merge(int u, int v) 30 { 31 if ((u = findFa(u)) == (v = findFa(v))) return; 32 if ((int)own[u].size() > (int)own[v].size()) swap(u, v); 33 fa[u] = v; 34 for (auto i : own[u]) own[v].insert(i); 35 own[u].clear(); 36 } 37 38 void addedge(int u, int v, int c) 39 { 40 if (mp[u].count(c)) merge(v, mp[u][c]); 41 if (mp[v].count(c)) merge(u, mp[v][c]); 42 mp[u][c] = v; mp[v][c] = u; 43 own[findFa(v)].insert(u); 44 own[findFa(u)].insert(v); 45 } 46 47 int main() 48 { 49 memset(fa, -1, sizeof(fa)); 50 scanf("%d %d %d %d ", &n, &m, &c, &q); 51 rep1(i, 1, m) 52 { 53 int u, v, c; scanf("%d %d %d ", &u, &v, &c); 54 u--, v--; 55 addedge(u, v, c); 56 } 57 while (q--) 58 { 59 char op; int x, y, z; 60 op = getchar(); 61 scanf("%d %d ", &x, &y); x--, y--; 62 if (op == '+') 63 { 64 scanf("%d ", &z); 65 addedge(x, y, z); 66 } 67 else 68 { 69 int rx = findFa(x); 70 if (rx == findFa(y) || own[rx].find(y) != own[rx].end()) 71 puts("Yes"); 72 else puts("No"); 73 } 74 } 75 return 0; 76 }
