本来在快乐写题的，突然A数据炸了，全场重测，unrated……
昨晚的题也有点毒，不过总体来说还算简单。
A：
一开始我也wa on 3了，仔细想想就会发现有重点的情况。
比如n==3, 3 1 2。答案应该是6而不是7。
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 fir first
9 #define sec second
10 #define sot(a,b) sort(a+1,a+1+b)
11 #define rep1(i,a,b) for(int i=a;i<=b;++i)
12 #define rep0(i,a,b) for(int i=a;i<b;++i)
13 #define repa(i,a) for(auto &i:a)
14 #define eps 1e-8
15 #define int_inf (1<<30)-1
16 #define ll_inf (1LL<<62)-1
17 #define lson curPos<<1
18 #define rson (curPos<<1)+1
19 /* namespace */
20 using namespace std;
21 /* header end */
22
23 int n, last = 0, ans = 0, flag = 1, llast = 0;
24
25 int main()
26 {
27 scanf("%d%d", &n, &last);
28 n--;
29 for (int i = 1; i <= n; i++)
30 {
31 int x;
32 scanf("%d", &x);
33 if (last == 1)
34 {
35 if (x == 1)
36 {
37 flag = 0;
38 }
39 else if (x == 2)
40 {
41 if (llast + last == 4)
42 ans += 2;
43 else
44 ans += 3;
45 }
46 else if (x == 3)
47 ans += 4;
48 }
49 else if (last == 2)
50 {
51 if (x == 2)
52 {
53 ans += 3;
54 }
55 else if (x == 1)
56 ans += 3;
57 else if (x == 3)
58 {
59 flag = 0;
60 }
61 }
62 else if (last == 3)
63 {
64 if (x == 3)
65 {
66 ans += 4;
67 }
68 else if (x == 1)
69 ans += 4;
70 else if (x == 2)
71 {
72 flag = 0;
73 }
74 }
75 if (i >= 1)
76 llast = last;
77 last = x;
78 }
79 if (!flag)
80 puts("Infinite");
81 else
82 {
83 puts("Finite");
84 printf("%d\n", ans);
85 }
86 return 0;
87 }
B：
各种写法都有。这题可以写得非常短。按字母在字母表的下标奇偶性来分类就好了。
然而我觉得取字母集合，sort一下再next_permutation+剪枝也可以，就是没有分类这种写法优秀。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 1e2 + 10;
6
7 void print(char *a, int len)
8 {
9 for (int i = 1; i <= len; i++) printf("%c", a[i]);
10 }
11
12 int main()
13 {
14 int t; scanf("%d", &t);
15 while (t--)
16 {
17 char s[maxn], a[maxn], b[maxn];
18 scanf("%s", s + 1); int len = strlen(s + 1), p = 0, q = 0;
19 sort(s + 1, s + 1 + len);
20 for (int i = 1; i <= len; i++)
21 if (s[i] & 1) a[++p] = s[i]; else b[++q] = s[i];
22 if (!p || !q)
23 {
24 print(a, p); print(b, q);
25 }
26 else if (abs(a[p] - b[1]) != 1)
27 {
28 print(a, p); print(b, q);
29 }
30 else if (abs(b[q] - a[1]) != 1)
31 {
32 print(b, q); print(a, p);
33 }
34 else printf("No answer");
35 puts("");
36 }
37 return 0;
38 }
C：
很容易就想到二分。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define int_inf (1<<30)-1
33 #define ll_inf (1LL<<62)-1
34 #define lson curPos<<1
35 #define rson (curPos<<1)+1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 const int maxn = 2e5 + 10;
41 int n, z, a[maxn];
42
43 int main()
44 {
45 scanf("%d%d", &n, &z);
46 for (int i = 0; i < n; i++) scanf("%d", &a[i]);
47 sort(a, a + n);
48 int l = 0, r = n / 2 + 1;
49 while (r - l > 1)
50 {
51 int mid = l + r >> 1, sign = 1;
52 for (int i = 0; i < mid; i++)
53 sign &= (a[n - mid + i] - a[i] >= z);
54 if (sign) l = mid; else r = mid;
55 }
56 printf("%d\n", l);
57 return 0;
58 }
D：
有点像是字典树，匹配文法规则为0*1*的字符串。
树上dp，也可以DSU过。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define int_inf 0x3f3f3f3f
33 #define ll_inf 0x7f7f7f7f7f7f7f7f
34 #define lson curPos<<1
35 #define rson curPos<<1|1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 const int maxn = 2e5 + 10;
41
42 int n, dp[2][maxn];
43 vector<pair<int, int>> graph[maxn];
44 ll ans;
45
46 void calc(int u, pair<int, int> v, int mt)
47 {
48 if (!v.second)
49 dp[0][u] += mt * dp[0][v.first];
50 else
51 dp[1][u] += mt * dp[0][v.first] + mt * dp[1][v.first];
52 }
53
54 void dfs1(int u, int p)
55 {
56 for (pair<int, int> v : graph[u])
57 {
58 if (v.first == p)
59 continue;
60 dfs1(v.first, u);
61 calc(u, v, 1);
62 }
63 ans += dp[0][u] + dp[1][u];
64 dp[0][u]++;
65 }
66
67 void dfs2(int u, int p, int pp1, int pp2)
68 {
69 ans += pp1 + pp2;
70 for (pair<int, int> v : graph[u])
71 {
72 if (v.first == p)
73 continue;
74 calc(u, v, -1);
75 if (!v.second)
76 dfs2(v.first, u, pp1 + dp[0][u], 0);
77 else
78 dfs2(v.first, u, 0, pp1 + pp2 + dp[0][u] + dp[1][u]);
79 calc(u, v, 1);
80 }
81 }
82
83 int main()
84 {
85 scanf("%d", &n);
86 for (int i = 1, u, v, c; i < n; i++)
87 {
88 scanf("%d%d%d", &u, &v, &c); --u; --v;
89 graph[u].pb(mp(v, c));
90 graph[v].pb(mp(u, c));
91 }
92 dfs1(0, -1); dfs2(0, -1, 0, 0);
93 cout << ans << endl;
94 return 0;
95 }
E：
看了rk1大佬的代码才意识到可以二分做。时间复杂度约等于O(NlogN)。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define int_inf (1<<30)-1
33 #define ll_inf (1LL<<62)-1
34 #define lson curPos<<1
35 #define rson (curPos<<1)+1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 const int maxn = 2e5 + 10;
41 int a[maxn], n;
42
43 ll solve(int l, int r)
44 {
45 if (r - l == 1) return 0;
46 int mid = l + r >> 1;
47 ll ret = solve(l, mid) + solve(mid, r); //避免处理长度为1的区间
48 {
49 set<int>s;
50 //把p当做区间的左端点&&最大值，往右扩展。下法同理。
51 for (int i = mid - 1, j = mid, p = 0; i >= l; i--)
52 {
53 p = max(p, a[i]);
54 while (j < r && a[j] < p) s.insert(a[j++]);
55 ret += s.count(p - a[i]);
56 }
57 }
58 {
59 set<int>s;
60 for (int i = mid, j = mid - 1, p = 0; i < r; i++)
61 {
62 p = max(p, a[i]);
63 while (j >= l && a[j] < p) s.insert(a[j--]);
64 ret += s.count(p - a[i]);
65 }
66 }
67 return ret;
68 }
69
70 int main()
71 {
72 scanf("%d", &n);
73 rep0(i, 0, n) scanf("%d", &a[i]);
74 printf("%lld\n", solve(0, n));
75 return 0;
76 }
F && G：
太神仙了……再见
原文地址：https://www.cnblogs.com/JHSeng/p/10803378.html
时间： 2024-10-17 07:31:36