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Codeforces Round #279 (Div. 2) 题解集合

时间:2014-11-23 23:09:24      阅读:398      评论:0      收藏:0      [点我收藏+]

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终于有场正常时间的比赛了。。。毛子换冬令时还正是好啊233

做了ABCD,E WA了3次最后没搞定,F不会= =

那就来说说做的题目吧= =

 

 A. Team Olympiad

水题嘛= =

就是个贪心什么的乱搞,貌似放A题难了

 

bubuko.com,布布扣
 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 const int N = 5005;
 6 
 7 int cnt[5], first[5], next[N];
 8 
 9 int main() {
10     int n ,i, x, ans;
11     int a, b, c;
12     scanf("%d", &n);
13     for (i = 1; i <= n; ++i) {
14         scanf("%d", &x);
15         next[i] = first[x], first[x] = i;
16         ++cnt[x];
17     }
18     ans = min(min(cnt[1], cnt[2]), cnt[3]);
19     printf("%d\n", ans);
20     a = first[1], b = first[2], c = first[3];
21     for (i = 1; i <= ans; ++i) {
22         printf("%d %d %d\n", a, b, c);
23         a = next[a], b = next[b], c = next[c];
24     }
25 }
View Code

 

B. Queue

这放B题真的合适吗= =

就是模拟啦,但是但是,具体处理好麻烦的说!!!

 

bubuko.com,布布扣
 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 const int N = (int) 1e6 + 5;
 6 int S = (int) 1e6 + 1;
 7 int T = (int) 1e6 + 2;
 8 
 9 struct edges {
10     int next, to;
11     edges() {}
12     edges(int _next, int _to) : next(_next), to(_to) {}
13 }e[N << 1];
14 
15 int n, tot, first[N];
16 int ans[N], cnt;
17 int Cnt[N];
18 bool v[N];
19 
20 inline void add_edges(int x, int y){
21     e[++tot] = edges(first[x], y), first[x] = tot;
22     e[++tot] = edges(first[y], x), first[y] = tot;    
23 }
24 
25 int main() {
26     int i, x, y;
27     scanf("%d", &n);
28     for (i = 1; i <= n; ++i) {
29         scanf("%d%d", &x, &y);
30         ++Cnt[x], --Cnt[y];
31         if (x == 0) x = S;
32         if (y == 0) y = T;
33         add_edges(x, y);
34     }
35     v[S] = 1, cnt = 1;
36     while (1) {
37         for (x = first[S]; x; x = e[x].next)
38             if (!v[e[x].to]) break;
39         if (x == 0) break;
40         v[S = e[x].to] = 1;
41         ans[cnt << 1] = S, ++cnt;
42     }
43     if (n & 1 == 0) {
44         v[T] = 1, cnt = 1;
45         while (1) {
46             for (x = first[T]; x; x = e[x].next)
47                 if (!v[e[x].to]) break;
48             if (x == 0) break;
49             v[T = e[x].to] = 1;
50             ans[n + 1 - (cnt << 1)] = T, ++cnt;
51         }
52     } else {
53         for (i = 1; i <= N; ++i)
54             if (Cnt[i] == 1) {
55                 S = i;
56                 break;
57             }
58         ans[1] = S;
59         v[S] = 1, cnt = 1;
60         while (1) {
61             for (x = first[S]; x; x = e[x].next)
62                 if (!v[e[x].to]) break;
63             if (x == 0) break;
64             v[S = e[x].to] = 1;
65             ans[cnt << 1 | 1] = S, ++cnt;
66         }
67     }
68     for (i = 1; i < n; ++i)
69         printf("%d ", ans[i]);
70     printf("%d\n", ans[n]);
71     return 0;
72 }
View Code

 

C. Hacking Cypher

正这反着扫两遍,直接判断就好了,报noip高精模写错的一箭之仇!

 

bubuko.com,布布扣
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 typedef long long ll;
 7 const int N = (int) 1e6 + 5;
 8 
 9 ll a, b;
10 int len;
11 bool ok_a[N], ok_b[N];
12 char st[N];
13 
14 int main() {
15     ll tmp, T;
16     int i, j;
17     scanf("%s", st + 1); len = strlen(st + 1);
18     scanf("%I64d%I64d", &a, &b);
19     tmp = 0;
20     for (i = 1; i <= len; ++i) {
21         ((tmp *= 10) += st[i] - 0 ) %= a;
22         if (tmp == 0) ok_a[i] = 1; else ok_a[i] = 0;
23     }
24     tmp = 0, T = 1;
25     for (i = len; i; --i) {
26         (tmp += (ll) T * (st[i] - 0)) %= b;
27         (T *= 10) %= b;
28         if (tmp == 0) ok_b[i] = 1; else ok_b[i] = 0;
29     }
30     for (i = 2; i <= len; ++i)
31         if (ok_a[i - 1] && ok_b[i] && st[i] != 0) break;
32     if (i == len + 1) {
33         puts("NO");
34         return 0;
35     }
36     puts("YES");
37     for (j = 1; j < i; ++j)
38         putchar(st[j]); puts("");
39     for (j = i; j <= len; ++j)
40         putchar(st[j]); puts("");
41     return 0;
42 }
View Code

 

D. Chocolate

第一眼神题Orz

后来发现,是指1 * 1的方格一样多。。。这尼玛是在逗我!

于是计算面积s1, s2,令s1 /= gcd, s2 /= gcd

然后判断s1 * (2 / 3) ^ x1 * (1 / 2) ^ y1 = s2 * (2 / 3) ^ x2 * (1 / 2) ^ y2 是否有非负整数解(x1, x2, y1, y2)且x1, x2; y1, y2中都至少有一个0

乱搞吧2333

 

bubuko.com,布布扣
 1 #include <cstdio>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 
 6 ll a, b, c, d, s1, s2, G;
 7 int ans;
 8 int cnt[2][2];
 9 
10 ll gcd(ll a, ll b) {
11     return !b ? a : gcd(b, a % b);
12 }
13 
14 int abs(int x) {
15     return x < 0 ? -x : x;
16 }
17 
18 void work() {
19     ans += cnt[0][1] + cnt[1][1] + abs(cnt[0][0] + cnt[0][1] - cnt[1][0] - cnt[1][1]);
20 }
21 
22 int main() {
23     scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);
24     s1 = a * b;
25     s2 = c * d;
26     G = gcd(s1, s2);
27     s1 /= G, s2 /= G;
28     while (!(s1 & 1)) s1 >>= 1, ++cnt[0][0];
29     while (s1 % 3 == 0) s1 /= 3, ++cnt[0][1];
30     while (!(s2 & 1)) s2 >>= 1, ++cnt[1][0];
31     while (s2 % 3 == 0) s2 /= 3, ++cnt[1][1];
32     if (s1 > 1 || s2 > 1) {
33         puts("-1");
34         return 0;
35     }
36     work();
37     printf("%d\n", ans);
38     cnt[0][0] += cnt[0][1], cnt[1][0] += cnt[1][1];
39     if (cnt[0][0] > cnt[1][0]) cnt[0][0] -= cnt[1][0], cnt[1][0] = 0;
40     else cnt[1][0] -= cnt[0][0], cnt[0][0] = 0;
41     while (cnt[0][1]--) {
42         if (a % 3 == 0) (a /= 3) *= 2;
43         else (b /= 3) *= 2;
44     }
45     while (cnt[1][1]--) {
46         if (c % 3 == 0) (c /= 3) *= 2;
47         else (d /= 3) *= 2;
48     }
49     while (cnt[0][0]--) {
50         if (!(a & 1)) a /= 2;
51         else b /= 2;
52     }
53     while (cnt[1][0]--) {
54         if (!(c & 1)) c /= 2;
55         else d /= 2;
56     }
57     printf("%I64d %I64d\n%I64d %I64d\n", a, b, c, d);
58     return 0;
59 }
View Code

 

E. Restoring Increasing Sequence

字符串处理一下,然后贪心当前最小即可,然后我的bin数组少了个0,WA到死啊!!!

我的Div.2 Rank 10-快还我。。。呜呜呜

 

bubuko.com,布布扣
 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 using namespace std;
 5 const int bin[9] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000};
 6 const int N = 100005;
 7 
 8 int n, len, len_last;
 9 char st[10];
10 int ans[N];
11 
12 int work(int p) {
13     int res = 0, i, j;
14     if (len_last > len) return 0;
15     if (len_last < len) {
16         for (i = 1; i <= len; ++i)
17             if (st[i] == ?)
18                 if (i == 1) res = 1; else res *= 10;
19             else (res *= 10) += st[i] - 0;
20         return res;
21     }
22     for (i = 1; i <= len; ++i)
23         if (st[i] == ?)
24             (res *= 10) += 9;
25         else (res *= 10) += st[i] - 0;
26     if (res <= ans[p - 1]) return 0;
27     for (i = 1; i <= len; ++i) 
28         if (st[i] == ?)
29             for (j = 1; j <= 9; ++j)
30                 if (res - bin[len - i] <= ans[p - 1]) break;
31                 else res -= bin[len - i];
32     return res;
33         
34 }
35 
36 int main() {
37     int i;
38     scanf("%d\n", &n);
39     ans[0] = 0;
40     len = 0;
41     for (i = 1; i <= n; ++i) {
42         scanf("%s\n", st + 1);
43         len_last = len, len = strlen(st + 1);
44         if (!(ans[i] = work(i))) {
45             puts("NO");
46             return 0;
47         }
48     }
49     puts("YES");
50     for (i = 1; i <= n; ++i)
51         printf("%d\n", ans[i]);
52     return 0;
53 }
View Code

 

F. Treeland Tour

第一反应是树上DP,每个点一个平衡树维护。。。

后来发现怎么可能,应该是点分治 + 归并数组。。。

但是真的能写的出来?不明= =(话说至今No Tags,什么东西!)

反正Div.2里只有一个人A了F,但是Div.1里A掉的貌似很多啊?以后再说吧

 

于是蒟蒻喜闻乐见的Div.2 Rank 44,被各位神犇D飞啦~Orz跪

Codeforces Round #279 (Div. 2) 题解集合

标签:style   blog   http   io   ar   color   os   sp   for   

原文地址:http://www.cnblogs.com/rausen/p/4117607.html

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