标签：开始 als class main lse clear ++i back char
A. The Fair Nut and Elevator
Solved.
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 110 5 int n, a[N]; 6 7 int main() 8 { 9 while (scanf("%d", &n) != EOF) 10 { 11 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 12 int res = 1e9; 13 for (int x = 1, tmp = 0; x <= 100; ++x, res = min(res, tmp), tmp = 0) for (int i = 1; i <= n; ++i) if (a[i]) 14 { 15 tmp += a[i] * (abs(x - i) + abs(i - 1) + abs(x - 1) + abs(x - 1) + abs(i - 1) + abs(x - i)); 16 } 17 printf("%d\n", res); 18 } 19 return 0; 20 }
B. Kvass and the Fair Nut
Upsolved.
题意：
刚开始题意读错，过了pretest就没管了
有$n桶啤酒，要从中取出s升，求取出后剩余的量最少的那桶啤酒最大$
思路：
二分或者直接算都行
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1010 6 int n; 7 ll s, v[N]; 8 9 bool check(ll x) 10 { 11 ll tot = 0; 12 for (int i = 1; i <= n; ++i) 13 { 14 if (v[i] < x) return false; 15 tot += v[i] - x; 16 } 17 return tot >= s; 18 } 19 20 int main() 21 { 22 while (scanf("%d%lld", &n, &s) != EOF) 23 { 24 for (int i = 1; i <= n; ++i) scanf("%lld", v + i); 25 ll l = 0, r = 1e9, res = -1; 26 while (l <= r) 27 { 28 ll mid = (l + r) >> 1; 29 if (check(mid)) 30 { 31 res = mid; 32 l = mid + 1; 33 } 34 else 35 r = mid - 1; 36 } 37 printf("%lld\n", res); 38 } 39 return 0; 40 }
C. The Fair Nut and String
Solved.
题意：
有一串字符串，求有多少个字符串，是abababab形式的。
思路：
字符不是‘a’ 也不是‘b’ 的话是没用的，然后如果两个‘a‘之间有多个‘b‘也没用
缩短字符串后
必然是aaabaaababab 这样的形式的
那么像统计DAG那样统计就没了
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 100010 6 const ll MOD = (ll)1e9 + 7; 7 char s[N], t[N]; 8 9 int main() 10 { 11 while (scanf("%s", s + 1) != EOF) 12 { 13 int n = 0; 14 for (int i = 1, len = strlen(s + 1); i <= len; ++i) if (s[i] == ‘a‘ || s[i] == ‘b‘) 15 { 16 if (s[i] == ‘a‘) t[++n] = ‘a‘; 17 else if (s[i] == ‘b‘ && t[n] == ‘a‘) t[++n] = ‘b‘; 18 } 19 if (n == 0) 20 { 21 puts("0"); 22 continue; 23 } 24 ll res = 1, pre = 1, cnt = 1; 25 for (int i = 2; i <= n; ++i) 26 { 27 if (t[i] == ‘b‘) 28 { 29 pre = (pre * (cnt + 1)) % MOD; 30 cnt = 0; 31 } 32 else 33 { 34 ++cnt; 35 res = (res + pre) % MOD; 36 } 37 } 38 printf("%lld\n", res); 39 } 40 return 0; 41 }
D. The Fair Nut and the Best Path
Upsolved.
题意：
在一棵树上，每个点是加油站，最多加油$w_i，然后每条边是路，耗费油v_i$
$定义f(u, v)为u->v的简单路径上每个点都加满油，假设油箱容量无限，最后剩下的油量$
如果其中某条路上油耗尽了，那么这条路是不可行的
思路：
我们把点权视为正直，边权视为负值
然后就是求任意两点之间的最大权值和
不需要考虑不合法的路径，因为如果存在不合法的路劲，
那么肯定存在另一条合法的路径使得答案比它更优。
令$f[u] 表示到达u的子树中某点的最大权， 这是纵向路径$
再考虑横向路径即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 300010 6 #define pii pair <int, int> 7 int n, w[N]; 8 vector <pii> G[N]; 9 ll res, f[N]; 10 11 void DFS(int u, int fa) 12 { 13 f[u] = w[u]; 14 ll Max[2] = {0, 0}; 15 for (auto it : G[u]) 16 { 17 int v = it.first; 18 if (v == fa) continue; 19 DFS(v, u); 20 ll cost = it.second; 21 f[u] = max(f[u], f[v] - cost + w[u]); 22 ll tmp = f[v] - cost; 23 if (tmp > Max[0]) 24 { 25 Max[1] = Max[0]; 26 Max[0] = tmp; 27 } 28 else if (tmp > Max[1]) 29 Max[1] = tmp; 30 } 31 res = max(res, max(f[u], Max[0] + Max[1] + w[u])); 32 } 33 34 int main() 35 { 36 while (scanf("%d", &n) != EOF) 37 { 38 for (int i = 1; i <= n; ++i) scanf("%d", w + i); 39 for (int i = 1; i <= n; ++i) G[i].clear(); 40 for (int i = 1, u, v, w; i < n; ++i) 41 { 42 scanf("%d%d%d", &u, &v, &w); 43 G[u].emplace_back(v, w); 44 G[v].emplace_back(u, w); 45 } 46 res = 0; 47 DFS(1, 1); 48 printf("%lld\n", res); 49 } 50 return 0; 51 }
E. The Fair Nut and Strings
Unsolved.
题意：
一共有k个长度为n的字符串，他们的范围是[s, t] 之间，按字典序排序
求这些字符串最多有多少个前缀
F. Max Mex
Unsolved.
Codeforces Round #526 (Div. 2) Solution
标签：开始 als class main lse clear ++i back char
原文地址：https://www.cnblogs.com/Dup4/p/10107846.html
