CodeCraft-19 and Codeforces Round #537 (Div. 2)
作者：互联网
A. Superhero Transformation
签
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1010
5 char s[N], t[N], Hash[N];
6
7 bool ok()
8 {
9 int n = strlen(s + 1);
10 for (int i = 1; i <= n; ++i)
11 if (Hash[s[i]] != Hash[t[i]])
12 return false;
13 return true;
14 }
15
16 int main()
17 {
18 memset(Hash, 0, sizeof Hash);
19 Hash['a'] = 1;
20 Hash['e'] = 1;
21 Hash['i'] = 1;
22 Hash['o'] = 1;
23 Hash['u'] = 1;
24 while (scanf("%s%s", s + 1, t + 1) != EOF)
25 {
26 int len1 = strlen(s + 1), len2 = strlen(t + 1);
27 if (len1 != len2) puts("No");
28 else
29 puts(ok() ? "Yes" : "No");
30 }
31 return 0;
32 }
View Code
B. Average Superhero Gang Power
Hacked.
注意如果最大值不止一个，那么去掉几个是不一定的
枚举即可。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define ll long long
5 #define N 100010
6 int n, k, m;
7 ll a[N];
8
9 int main()
10 {
11 while (scanf("%d%d%d", &n, &k, &m) != EOF)
12 {
13 for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
14 sort(a + 1, a + 1 + n);
15 for (int i = 1; i <= n; ++i) a[i] += a[i - 1];
16 double res = a[n] * 1.0 / n;
17 for (int i = 0; i < n; ++i)
18 {
19 if (m < i) break;
20 ll add = min(1ll * (m - i), 1ll * k * (n - i));
21 res = max(res, (a[n] - a[i] + add) * 1.0 / (n - i));
22 }
23 printf("%.10f\n", res);
24 }
25 return 0;
26 }
View Code
C. Creative Snap
递归求解即可，注意一整段空直接返回$A$
这样最多只有$k个长度为1的节点被访问到$
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define ll long long
5 #define N
6 int n, k, A, B;
7 vector <int> v;
8
9 ll CDQ(ll l, ll r)
10 {
11 if (r < l) return 0;
12 int num = upper_bound(v.begin(), v.end(), r) - lower_bound(v.begin(), v.end(), l);
13 if (num == 0) return A;
14 ll tmp = 1ll * (r - l + 1) * B * num;
15 if (l == r) return tmp;
16 ll mid = (l + r) >> 1;
17 return min(tmp, CDQ(l, mid) + CDQ(mid + 1, r));
18 }
19
20 int main()
21 {
22 while (scanf("%d%d%d%d", &n, &k, &A, &B) != EOF)
23 {
24 v.clear();
25 for (int i = 1, x; i <= k; ++i)
26 {
27 scanf("%d", &x);
28 v.push_back(x);
29 }
30 sort(v.begin(), v.end());
31 printf("%lld\n", CDQ(1, 1 << n));
32 }
33 return 0;
34 }
View Code
D. Destroy the Colony
Solved.
题意：
有$n个人，n为偶数, 每个人有一个种类$
$询问给出x, y 要求将和第x人同种类并且和第y个人同种类的人放在一个集合$
$并且挑选出一些人和他们放在一起，剩下的人在另一个集合$
$要求两个集合人数相等，并且同一种类的人属于同个集合$
$给出有多少种分配方式, 集合里是有顺序的,并且两个集合是不同的$
思路：
考虑一共有$f种方式将其他人和第x人同种类的人以及第y人同种类的人放在一个集合$
$我们假定 第i种类的人有c_i个, 令m = \frac{n}{2}$
$那么答案就是 \frac {m! \cdot m! \cdot f \cdot 2}{c_1! \cdot c_2! \cdots}$
那么考虑怎么处理$f, 可以用背包，但是暴力预处理时间复杂度是O(52^{3} \cdot n)$
$但是我们可以先处理好总的，注意到对于每种情况的处理只有最多两项物品的删除$
$直接删除方案数即可 这样时间复杂度就是O(52^{2} \cdot n)$
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];
8 int cnt[110], Hash[210];
9 int n, q, x, y, mid;
10 ll fac[N], inv[N];
11 ll f[110][110];
12 ll g[N];
13
14 ll qmod(ll base, ll n)
15 {
16 ll res = 1;
17 while (n)
18 {
19 if (n & 1) res = res * base % MOD;
20 base = base * base % MOD;
21 n >>= 1;
22 }
23 return res;
24 }
25
26 ll C(int n, int m)
27 {
28 return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
29 }
30
31 int main()
32 {
33 fac[0] = 1;
34 for (int i = 1; i <= 100000; ++i) fac[i] = (fac[i - 1] * i) % MOD;
35 inv[100000] = qmod(fac[100000], MOD - 2);
36 for (int i = 100000; i >= 1; --i) inv[i - 1] = (inv[i] * i) % MOD;
37 for (int i = 1; i <= 26; ++i) Hash['a' + i - 1] = i;
38 for (int i = 27; i <= 52; ++i) Hash['A' + i - 27] = i;
39 while (scanf("%s", s + 1) != EOF)
40 {
41 n = strlen(s + 1);
42 mid = n / 2;
43 memset(cnt, 0, sizeof cnt);
44 for (int i = 1; i <= n; ++i) ++cnt[Hash[s[i]]];
45 ll tmp = 1;
46 for (int i = 1; i <= 52; ++i)
47 if (cnt[i]) tmp = (tmp * inv[cnt[i]]) % MOD;
48 memset(f, 0, sizeof f);
49 memset(g, 0, sizeof g);
50 g[0] = 1;
51 for (int i = 1; i <= 52; ++i) if (cnt[i])
52 {
53 for (int j = mid; j >= cnt[i]; --j)
54 g[j] += g[j - cnt[i]];
55 }
56 for (int i = 1; i <= 52; ++i)
57 {
58 for (int j = i + 1; j <= 52; ++j)
59 {
60 if (cnt[i] == 0 || cnt[j] == 0) continue;
61 if (cnt[i] + cnt[j] > mid) continue;
62 if (cnt[i] + cnt[j] == mid)
63 {
64 f[i][j] = f[j][i] = 1;
65 continue;
66 }
67 for (int k = cnt[i]; k <= mid; ++k)
68 g[k] -= g[k - cnt[i]];
69 for (int k = cnt[j]; k <= mid; ++k)
70 g[k] -= g[k - cnt[j]];
71 f[i][j] = f[j][i] = g[mid - cnt[i] - cnt[j]] % MOD;
72 //printf("%d %d %lld\n", i, j, f[i][j]);
73 for (int k = mid; k >= cnt[i]; --k)
74 g[k] += g[k - cnt[i]];
75 for (int k = mid; k >= cnt[j]; --k)
76 g[k] += g[k - cnt[j]];
77 }
78 }
79 for (int i = 1; i <= 52; ++i) if (cnt[i] && cnt[i] <= mid)
80 {
81 for (int j = cnt[i]; j <= mid; ++j)
82 g[j] -= g[j - cnt[i]];
83 f[i][i] = g[mid - cnt[i]] % MOD;
84 //printf("%d %lld\n", i, f[i][i]);
85 //printf("%d %d %lld\n", i, i, f[i][i]);
86 for (int j = mid; j >= cnt[i]; --j)
87 g[j] += g[j - cnt[i]];
88 }
89 //for (int i = 1; i <= mid; ++i) printf("%lld%c", g[i], " \n"[i == mid]);
90 scanf("%d", &q);
91 while (q--)
92 {
93 scanf("%d%d", &x, &y);
94 //cout << f[Hash[s[x]]][Hash[s[y]]] << endl;
95 ll res = fac[mid] * fac[mid] % MOD * 2 % MOD * tmp % MOD * f[Hash[s[x]]][Hash[s[y]]] % MOD;
96 printf("%lld\n", res);
97 }
98 }
99 return 0;
100 }
View Code
标签：CodeCraft,return,19,ll,Codeforces,mid,int,cnt,define 来源： https://www.cnblogs.com/Dup4/p/10351424.html