Codeforces Round #279 (Div. 2) 题解集合

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

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = ;

 int cnt[], first[], next[N];

 int main() {
     int n ,i, x, ans;
     int a, b, c;
     scanf("%d", &n);
     for (i = ; i <= n; ++i) {
         scanf("%d", &x);
         next[i] = first[x], first[x] = i;
         ++cnt[x];
     }
     ans = min(min(cnt[], cnt[]), cnt[]);
     printf("%d\n", ans);
     a = first[], b = first[], c = first[];
     for (i = ; i <= ans; ++i) {
         printf("%d %d %d\n", a, b, c);
         a = next[a], b = next[b], c = next[c];
     }
 }

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！！！

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = (int) 1e6 + ;
 int S = (int) 1e6 + ;
 int T = (int) 1e6 + ;

 struct edges {
     int next, to;
     edges() {}
     edges(int _next, int _to) : next(_next), to(_to) {}
 }e[N << ];

 int n, tot, first[N];
 int ans[N], cnt;
 int Cnt[N];
 bool v[N];

 inline void add_edges(int x, int y){
     e[++tot] = edges(first[x], y), first[x] = tot;
     e[++tot] = edges(first[y], x), first[y] = tot;
 }

 int main() {
     int i, x, y;
     scanf("%d", &n);
     for (i = ; i <= n; ++i) {
         scanf("%d%d", &x, &y);
         ++Cnt[x], --Cnt[y];
         if (x == ) x = S;
         if (y == ) y = T;
         add_edges(x, y);
     }
     v[S] = , cnt = ;
     while () {
         for (x = first[S]; x; x = e[x].next)
             if (!v[e[x].to]) break;
         if (x == ) break;
         v[S = e[x].to] = ;
         ans[cnt << ] = S, ++cnt;
     }
     if (n &  == ) {
         v[T] = , cnt = ;
         while () {
             for (x = first[T]; x; x = e[x].next)
                 if (!v[e[x].to]) break;
             if (x == ) break;
             v[T = e[x].to] = ;
             ans[n +  - (cnt << )] = T, ++cnt;
         }
     } else {
         for (i = ; i <= N; ++i)
             if (Cnt[i] == ) {
                 S = i;
                 break;
             }
         ans[] = S;
         v[S] = , cnt = ;
         while () {
             for (x = first[S]; x; x = e[x].next)
                 if (!v[e[x].to]) break;
             if (x == ) break;
             v[S = e[x].to] = ;
             ans[cnt <<  | ] = S, ++cnt;
         }
     }
     for (i = ; i < n; ++i)
         printf("%d ", ans[i]);
     printf("%d\n", ans[n]);
     return ;
 }

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！

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 typedef long long ll;
 const int N = (int) 1e6 + ;

 ll a, b;
 int len;
 bool ok_a[N], ok_b[N];
 char st[N];

 int main() {
     ll tmp, T;
     int i, j;
     scanf("%s", st + ); len = strlen(st + );
     scanf("%I64d%I64d", &a, &b);
     tmp = ;
     for (i = ; i <= len; ++i) {
         ((tmp *= ) += st[i] - '' ) %= a;
         if (tmp == ) ok_a[i] = ; else ok_a[i] = ;
     }
     tmp = , T = ;
     for (i = len; i; --i) {
         (tmp += (ll) T * (st[i] - '')) %= b;
         (T *= ) %= b;
         if (tmp == ) ok_b[i] = ; else ok_b[i] = ;
     }
     for (i = ; i <= len; ++i)
         if (ok_a[i - ] && ok_b[i] && st[i] != '') break;
     if (i == len + ) {
         puts("NO");
         return ;
     }
     puts("YES");
     for (j = ; j < i; ++j)
         putchar(st[j]); puts("");
     for (j = i; j <= len; ++j)
         putchar(st[j]); puts("");
     return ;
 }

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

 #include <cstdio>

 using namespace std;
 typedef long long ll;

 ll a, b, c, d, s1, s2, G;
 int ans;
 int cnt[][];

 ll gcd(ll a, ll b) {
     return !b ? a : gcd(b, a % b);
 }

 int abs(int x) {
     return x <  ? -x : x;
 }

 void work() {
     ans += cnt[][] + cnt[][] + abs(cnt[][] + cnt[][] - cnt[][] - cnt[][]);
 }

 int main() {
     scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);
     s1 = a * b;
     s2 = c * d;
     G = gcd(s1, s2);
     s1 /= G, s2 /= G;
     while (!(s1 & )) s1 >>= , ++cnt[][];
     while (s1 %  == ) s1 /= , ++cnt[][];
     while (!(s2 & )) s2 >>= , ++cnt[][];
     while (s2 %  == ) s2 /= , ++cnt[][];
     if (s1 >  || s2 > ) {
         puts("-1");
         return ;
     }
     work();
     printf("%d\n", ans);
     cnt[][] += cnt[][], cnt[][] += cnt[][];
     if (cnt[][] > cnt[][]) cnt[][] -= cnt[][], cnt[][] = ;
     else cnt[][] -= cnt[][], cnt[][] = ;
     while (cnt[][]--) {
         if (a %  == ) (a /= ) *= ;
         else (b /= ) *= ;
     }
     while (cnt[][]--) {
         if (c %  == ) (c /= ) *= ;
         else (d /= ) *= ;
     }
     while (cnt[][]--) {
         if (!(a & )) a /= ;
         else b /= ;
     }
     while (cnt[][]--) {
         if (!(c & )) c /= ;
         else d /= ;
     }
     printf("%I64d %I64d\n%I64d %I64d\n", a, b, c, d);
     return ;
 }

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

 #include <cstdio>
 #include <cstring>

 using namespace std;
 const int bin[] = {, , , , , , , , };
 const int N = ;

 int n, len, len_last;
 char st[];
 int ans[N];

 int work(int p) {
     int res = , i, j;
     if (len_last > len) return ;
     if (len_last < len) {
         for (i = ; i <= len; ++i)
             if (st[i] == '?')
                 if (i == ) res = ; else res *= ;
             else (res *= ) += st[i] - '';
         return res;
     }
     for (i = ; i <= len; ++i)
         if (st[i] == '?')
             (res *= ) += ;
         else (res *= ) += st[i] - '';
     if (res <= ans[p - ]) return ;
     for (i = ; i <= len; ++i)
         if (st[i] == '?')
             for (j = ; j <= ; ++j)
                 if (res - bin[len - i] <= ans[p - ]) break;
                 else res -= bin[len - i];
     return res;

 }

 int main() {
     int i;
     scanf("%d\n", &n);
     ans[] = ;
     len = ;
     for (i = ; i <= n; ++i) {
         scanf("%s\n", st + );
         len_last = len, len = strlen(st + );
         if (!(ans[i] = work(i))) {
             puts("NO");
             return ;
         }
     }
     puts("YES");
     for (i = ; i <= n; ++i)
         printf("%d\n", ans[i]);
     return ;
 }

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

话说，蒟蒻的Rating曲线越来越了难看了233