没想到平成年代最后一场cf居然是手速场，幸好拿了个小号娱乐不然掉分预定 (逃
A：
傻题。
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 = 1e2 + 10;
41 int n, m, r, minn = 2000, maxx = 0;
42
43 int main()
44 {
45 scanf("%d%d%d", &n, &m, &r);
46 rep1(i, 1, n)
47 {
48 int x; scanf("%d", &x); minn = min(minn, x);
49 }
50 rep1(i, 1, m)
51 {
52 int x; scanf("%d", &x); maxx = max(maxx, x);
53 }
54 if (maxx > minn) printf("%d\n", maxx * (r / minn) + r % minn);
55 else printf("%d\n", r);
56 return 0;
57 }
B：
乍一看以为要搜，其实根本不需要，n^2填进去判断就完事了 (看通过人数就能猜到根本不用搜。
至于为什么可以这样，是因为跟紧密填充相关吗？
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 = 55;
41 int a[maxn][maxn], n, solved = 0;
42 char s[maxn];
43
44 int check()
45 {
46 rep1(i, 1, n)
47 rep1(j, 1, n)
48 if (!a[i][j]) return 0;
49 return 1;
50 }
51
52 int putable(int x, int y)
53 {
54 if (x == 1 || x == n || y == 1 || y == n) return 0;
55 if (a[x - 1][y] || a[x + 1][y] || a[x][y - 1] || a[x][y + 1] || a[x][y]) return 0;
56 return 1;
57 }
58
59 int main()
60 {
61 scanf("%d", &n);
62 rep1(i, 1, n)
63 {
64 scanf("%s", s + 1);
65 rep1(j, 1, n)
66 if (s[j] == ‘#‘) a[i][j] = 1; else a[i][j] = 0;
67 }
68 rep1(i, 2, n - 1)
69 {
70 rep1(j, 2, n - 1)
71 if (putable(i, j))
72 {
73 a[i - 1][j] = a[i + 1][j] = a[i][j - 1] = a[i][j + 1] = a[i][j] = 1;
74 }
75 }
76 if (check()) puts("YES");
77 else puts("NO");
78 return 0;
79 }
C：
2e5质数筛贪心就完事了，优先满足最近的质数，先塞2再塞1，也有大佬说不用筛。
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 int c1 = 0, c2 = 0, n, prime[maxn], p = 1, tot;
42 vector<int>ans;
43
44 bool valid[maxn];
45 void getPrime(int n, int &tot, int ans[maxn])
46 {
47 tot = 0;
48 memset(valid, 1, sizeof(valid));
49 for (int i = 2; i <= n; i++)
50 {
51 if (valid[i]) ans[++tot] = i;
52 for (int j = 1; (j <= tot) && (i * ans[j] <= n); j++)
53 {
54 valid[i * ans[j]] = false;
55 if (i % ans[j] == 0) break;
56 }
57 }
58 }
59
60 int main()
61 {
62 getPrime(2e5, tot, prime);
63 ans.clear();
64 scanf("%d", &n);
65 rep1(i, 1, n)
66 {
67 int x; scanf("%d", &x);
68 if (x & 1) c1++; else c2++;
69 }
70 int curr = 0;
71 while (c1 || c2)
72 {
73 if (prime[p] - curr > 2 && c2)
74 {
75 ans.pb(2); c2--; curr += 2;
76 }
77 else if (prime[p] - curr == 2 && c2)
78 {
79 ans.pb(2); c2--; p++; curr += 2;
80 }
81 else if (prime[p] - curr == 1 && c1)
82 {
83 ans.pb(1); c1--; p++; curr += 1;
84 }
85 else
86 {
87 if (c2)
88 {
89 ans.pb(2); c2--; curr += 2;
90 }
91 else if (c1)
92 {
93 ans.pb(1); c1--; curr += 1;
94 }
95 }
96 }
97 for (auto i : ans) printf("%d ", i);
98 return 0;
99 }
D：
把全场人卡死的dp，直到比赛结束都只有不到20个人过了，待补。
E：
看了一眼感觉其实比D简单(我口胡的)，待补。
原文地址：https://www.cnblogs.com/JHSeng/p/10795255.html
时间： 2024-10-08 03:07:11