NOIP模拟测试21

Problem A:折纸

为了好做一点,直接钦定往左折,折完如果非法就调整一下(

可以开一个坐标数组记录每个点折叠后的坐标zzz 60pts

然后发现边界只有可能是处理过的点,就不用每个点都记了,只记处理过的点就行了,100pts

 1 #include <bits/stdc++.h>
 2 
 3 int m;
 4 long long n;
 5 long long sega[3005];
 6 
 7 signed main() {
 8     scanf("%lld%d", &n, &m);
 9     long long l = 0, r = n;
10     for (int i = 1; i <= m; i++) {
11         scanf("%lld", &sega[i]);
12         for (int j = 1; j <= i - 1; j++) {
13             if (sega[j] < sega[i])
14                 sega[i] = 2 * sega[j] - sega[i];
15         }
16         l = std::min(l, 2 * sega[i] - r), r = sega[i];
17         if (l > r) std::swap(l, r);
18     }
19     printf("%lld\n", r - l);
20     return 0;
21 }
A

Problem B:不等式

当没有%M我们裆燃会做啦

$L \le Sx \le R$ 用L求个x,判断Sx是否小于R。

当拓展到%M,我们改写一下:$L \le Sx - My \le R$.

移项乱搞后:$-R \mod S \le My \le -L \mod S$

可以继续递归下去(

再把边界搞一下就vans了

 1 #include <bits/stdc++.h>
 2 #define bu_kai_long return
 3 #define long_jian_zu_zong 0
 4 
 5 int T;
 6 long long m, s, l, r;
 7 
 8 long long solve(long long m, long long s, long long l, long long r) {
 9     if (l >= m || l > r) return -1;
10     if (l == 0) return 0;
11     s %= m;
12     if (s == 0) return -1;
13     long long ans = (l - 1) / s + 1;
14     if (ans * s <= r && ans * s < m) return ans;
15     long long L = (-r % s + s) % s, R = (-l % s + s) % s;
16     long long misaka = solve(s, m, L, R);
17     if (misaka == -1) return -1;
18     long long mikoto = (l + m * misaka - 1) / s + 1;
19     if (s * mikoto - m * misaka <= r) return mikoto;
20     return -1;
21 }
22 
23 signed main() {
24     scanf("%d", &T);
25     while (T--) {
26         scanf("%lld%lld%lld%lld", &m, &s, &l, &r);
27         if (r >= m) r = m - 1;
28         printf("%lld\n", solve(m, s, l, r));
29     }
30     bu_kai_long long_jian_zu_zong;
31 }
B

Problem C:reverse

 是我写不出来的数位DP:)

考场40分性价比还是很高的(

正解:设f[i][j][op1][op2]表示前i位,当前后缀长度为j,前缀reverse后与L比较的结果和比R比较的结果

这没啥说的,记搜就vans了

 1 #include <bits/stdc++.h>
 2 #define ull unsigned long long
 3 
 4 int T, a, b;
 5 ull l, r;
 6 
 7 namespace subtask1 {
 8     void QAQ() {
 9         printf("%llu\n", r);
10     }
11 }
12 
13 namespace subtask3 {
14     int A[23], B[23], C[23];
15     ull f[23][23][4][4];
16     bool vis[23][23][4][4];
17 
18     int cmp(int x, int y, int las) {
19         if (x < y) return 0;
20         else if (x == y) return las;
21         else return 2;
22     }
23 
24     ull dfs(int pos, bool lim, int len, int op1, int op2) {
25         if (!pos) {
26             if (len < A[0]) op1 = 0;
27             if (len < B[0]) op2 = 0;
28             return (op1 && op2 != 2) ? 1 : 0;
29         } 
30         if (!lim && vis[pos][len][op1][op2]) return f[pos][len][op1][op2];
31         int upp = lim ? C[pos] : 9;
32         ull ans = 0;
33         for (int i = 0; i <= upp; i++) {
34             ans += dfs(pos - 1, lim && i == upp, 
35                     len + 1, cmp(i, A[len + 1], op1), cmp(i, B[len + 1], op2));
36         }
37         if (!lim) {
38             vis[pos][len][op1][op2] = 1;
39             f[pos][len][op1][op2] = ans;
40         }
41         return ans;
42     }
43 
44     void seprate(ull x, int dig[]) {
45         dig[0] = 0;
46         while (x)
47             dig[++dig[0]] = x % 10, x /= 10;
48     }
49 
50     ull yukari() {
51         memset(vis, 0, sizeof(vis));
52         memset(f, 0, sizeof(f));
53         ull ret = 0;
54         for (int i = 1; i <= C[0]; i++) {
55             int upp = i == C[0] ? C[C[0]] : 9;
56             for (int j = 1; j <= upp; j++) {
57                 ret += dfs(i - 1, j == upp && i == C[0], 
58                         1, cmp(j, A[1], 1), cmp(j, B[1], 1));    
59             }
60         }
61         return ret;
62     }
63 
64     void QAQ() {
65         memset(A, 0, sizeof(A));
66         memset(B, 0, sizeof(B));
67         memset(C, 0, sizeof(C));
68         seprate(l, A), seprate(r, B), seprate(r, C);
69         ull ans = yukari();
70         memset(C, 0, sizeof(C));
71         seprate(l - 1, C);
72         ans -= yukari();
73         printf("%llu\n", ans);
74     }
75 }
76 
77 signed main() {
78     scanf("%d%d%d", &T, &a, &b);
79     while (T--) {
80         scanf("%llu%llu", &l, &r);
81         if (a && b) subtask1::QAQ();
82         else subtask3::QAQ();
83     }
84     return 0;
85 }
C

 

posted @ 2019-08-18 19:25  Gekoo  阅读(...)  评论(...编辑  收藏