[codeforces#592Div2]C-G题

题目链接

感觉这场难度迷茫,个人觉得难度排序为$A<B<D=E=G<C<F$


 

C题:

比赛结束1500+pp,结果出分900+fst,我就是fst的睿智Orz。

题意为给出$n,p,w,d$,求满足下列式子的任意$x,y,z$

$x*w+y*d=p\&\& x+y+z=n\&\&x\geq 0\&\&y\geq 0\&\&z\geq 0$

如果不看$z$,式子的前半段就是扩展欧几里得,所以先求出式子$x*w+y*d=p$的一种解$(x,y)$,然后再判断解的合法性,并将解转化成合法的。

由于求解时会爆longlong,所以用的java。

 1 import java.math.BigInteger;
 2 import java.util.*;
 3 public class Main {
 4     public static BigInteger x,y;
 5     public static void main(String[] args) {
 6         Scanner in = new Scanner(System.in);
 7         BigInteger n,p,w,d,gcd,xx,yy;
 8         n = in.nextBigInteger();
 9         p = in.nextBigInteger();
10         w = in.nextBigInteger();
11         d = in.nextBigInteger();
12         gcd = exgcd(w,d);
13         if(p.mod(gcd)!=BigInteger.ZERO) {
14             System.out.printf("-1\n");
15             return ;
16         }
17         x=x.multiply(p.divide(gcd));
18         y=y.multiply(p.divide(gcd));
19         BigInteger lcm = w.divide(gcd).multiply(d);
20         xx = lcm.divide(w);
21         yy = lcm.divide(d);
22         if (x.compareTo(BigInteger.ZERO)==-1 &&y.compareTo(BigInteger.ZERO)==-1) {
23             System.out.printf("-1\n");
24             return ;
25         }
26         if (x.compareTo(BigInteger.ZERO)==-1) {
27             x=x.add (y.divide(yy).multiply(xx));
28             y=y.mod(yy);
29         }
30         else if (y.compareTo(BigInteger.ZERO)==-1) {
31             y=y.add (x.divide(xx).multiply(yy));
32             x=x.mod(xx);
33         }
34         if (x.add(y).compareTo(n)!=1 && x.compareTo(BigInteger.ZERO)!=-1 && y .compareTo(BigInteger.ZERO)!=-1) {
35             System.out.print(x);
36             System.out.print(" ");
37             System.out.print(y);
38             System.out.print(" ");
39             System.out.print(n.subtract(x.add(y)));
40             return ;
41         }
42         x=x.add (y.divide(yy).multiply(xx));
43         y=y.mod(yy);
44         if (x.add(y).compareTo(n)!=1 && x.compareTo(BigInteger.ZERO)!=-1 && y.compareTo(BigInteger.ZERO)!=-1&&x.multiply(w).add(y.multiply(d)).compareTo(p)==0) {
45             System.out.print(x);
46             System.out.print(" ");
47             System.out.print(y);
48             System.out.print(" ");
49             System.out.print(n.subtract(x.add(y)));
50             return ;
51         }
52         else {
53             System.out.printf("-1\n");
54             return ;
55         }
56     }
57     public static BigInteger exgcd(BigInteger a,BigInteger b) {
58         if (b == BigInteger.ZERO) {
59             x = BigInteger.ONE;
60             y = BigInteger.ZERO;
61             return a;
62         }
63         BigInteger g = exgcd(b, a.mod(b));
64         BigInteger t = x;
65         x = y;
66         y =t.subtract((a.divide(b)).multiply(y));
67         return g;
68     }
69 }

D题:

题意是给一棵树每个点染色,一共三种颜色,求染色花费最少并且相邻的三个点颜色不能重复。

因为相邻的三个点颜色不能一样,所以树上每个点的度都要$\leq 2$,即只有链才能染色。而链上染色只有$6$种方法,所以枚举$6$次即可。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5 + 10;
 5 const ll inf = 1e18;
 6 ll a[maxn][5], ans[maxn];
 7 vector<int>p[maxn];
 8 int cnt[10][3] = { {1,2,3},{2,1,3},{3,1,2},{1,3,2},{2,3,1},{3,2,1} };
 9 ll dfs(int x, int fa, int w, int t) {
10     ll ans = a[x][cnt[w][t]];
11     for (int i = 0; i < p[x].size(); i++) {
12         int y = p[x][i];
13         if (y == fa)continue;
14         ans += dfs(y, x, w, (t + 1) % 3);
15     }
16     return ans;
17 }
18 void dfs1(int x, int fa, int w, int t) {
19     ans[x] = cnt[w][t];
20     for (int i = 0; i < p[x].size(); i++) {
21         int y = p[x][i];
22         if (y == fa)continue;
23         dfs1(y, x, w, (t + 1) % 3);
24     }
25 }
26 int main() {
27     int n;
28     scanf("%d", &n);
29     for (int j = 1; j <= 3; j++)
30         for (int i = 1; i <= n; i++)
31             scanf("%lld", &a[i][j]);
32     int f = 0, root = 1;
33     for (int i = 1, x, y; i < n; i++) {
34         scanf("%d%d", &x, &y);
35         p[x].push_back(y);
36         p[y].push_back(x);
37         if (p[x].size() > 2 || p[y].size() > 2)
38             f = 1;
39     }
40     for (int i = 1; i <= n; i++)
41         if (p[i].size() == 1)
42             root = i;
43     if (f == 1)
44         printf("-1\n");
45     else {
46         ll Min = inf, ansi = 0;
47         for (int i = 0; i < 6; i++) {
48             ll t = dfs(root, 0, i, 0);
49             if (Min > t)
50                 Min = t, ansi = i;
51         }
52         printf("%lld\n", Min);
53         dfs1(root, 0, ansi, 0);
54         for (int i = 1; i <= n; i++)
55             printf("%lld%c", ans[i], i == n ? '\n' : ' ');
56     }
57 }

E题:

题意说有$n$个数,有一种操作可以让一个数$+1$或$-1$,问最多$k$次操作后$min(max(a_{i})-min(a_{i}))$的值。

排序后记录一下每种数的个数,然后从首尾扫,判断哪种数的个数少,然后就乱搞。感觉比之前的要简单...

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5 + 10;
 5 ll a[maxn];
 6 pair<ll, ll>b[maxn];
 7 int main() {
 8     ll n, k, cnt = 0, num = 0;
 9     scanf("%lld%lld", &n, &k);
10     for (int i = 1; i <= n; i++)
11         scanf("%lld", &a[i]);
12     sort(a + 1, a + 1 + n);
13     for (int i = 1; i <= n + 1; i++) {
14         if (a[i - 1] && a[i] != a[i - 1])
15             b[++cnt] = pair<ll, ll>(a[i - 1], num), num = 0;
16         num++;
17     }
18     ll l = 1, r = cnt;
19     while (k && l != r) {
20         if (b[l].second <= b[r].second) {
21             int cg = min(b[l + 1].first - b[l].first, k / b[l].second);
22             if (cg == 0)break;
23             b[l].first += cg;
24             k -= b[l].second * cg;
25             if (b[l].first == b[l + 1].first)b[l + 1].second += b[l].second, l++;
26         }
27         else {
28             int cg = min(b[r].first - b[r-1].first, k / b[r].second);
29             if (cg == 0)break;
30             b[r].first -= cg;
31             k -= b[r].second * cg;
32             if (b[r].first == b[r-1].first)b[r - 1].second += b[r].second, r--;
33         }
34     }
35     printf("%lld\n", b[r].first - b[l].first);
36 }

F题:

题意是说一个有n个黑白点的环,操作k次,每次操作时对于每个点,点$i$的颜色变为$i,i-1,i+1$三个点颜色的众数,即为点$i-1$为黑,点$i+1$为黑,点$i$为白,则点$i$为黑。

大致画几个图就会发现,如果有两个即以上的相邻点同色,则块内的点的颜色永远都不会被改变。

所以我们只要找到每个点到达最终状态需要多少次就可以得到,将所有在块内的点初始为0,其余的状态扫正反扫两遍可以得到。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e5 + 10;
 5 char s[200010];
 6 int vis[200010];
 7 int main() {
 8     memset(vis, 0x3f, sizeof(vis));
 9     int n, k;
10     scanf("%d%d", &n, &k);
11     scanf("%s", s);
12     for (int i = 0; i < n; i++)
13         if (s[i] == s[(i + 1) % n] || s[i] == s[(i - 1 + n) % n])
14             vis[i] = 0;
15     for (int i = 2 * n - 1; i >= 0; i--)
16         vis[i % n] = min(vis[i % n], vis[(i + 1) % n] + 1);
17     for (int i = 0; i < 2 * n; i++)
18         vis[i % n] = min(vis[i % n], vis[(i - 1 + n) % n] + 1);
19     for (int i = 0; i < n; i++)
20         if (min(vis[i], k) % 2 == 1)
21             printf("%c", 'W' + 'B' - s[i]);
22         else
23             printf("%c", s[i]);
24     return 0;
25 }

 


G题:

题意是说有两个队列$q,p$,每个队列有$n$个人,编号为$1-n$,求最大的$ans=\sum_{i=1}^{n}max(p_{i},q_{i})\&\&ans\leq k$

先固定一个序列$q$,为$1,2,3\cdot \cdot \cdot n$,然后去构造序列$p$。

序列$p$初始也为$1,2,3\cdot \cdot \cdot n$,如果交换$p_{1},p_{n}$后的答案小于等于$k$,则交换,然后判断能否交换$p_{2},p_{n-1}$。

如果交换$p_{1},p_{n}$后的答案大于$k$,则直接找到位置$x$,使得交换$p_{1},p_{x}$后答案等于k。依次类推。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e6 + 10;
 5 int ans[maxn];
 6 int main() {
 7     ll n, k;
 8     scanf("%lld%lld", &n, &k);
 9     ll sum = (n + 1) * n >> 1;
10     if (k < sum) {
11         printf("-1\n");
12         return 0;
13     }
14     for (int i = 1; i <= n; i++)
15         ans[i] = i;
16     int l = 1, r = n;
17     while (l < r) {
18         if (sum + r - l <= k) {
19             swap(ans[l], ans[r]);
20             sum += r - l; l++; r--;
21         }
22         else {
23             int p = k - sum + l;
24             swap(ans[l], ans[p]);
25             sum += p - l;
26             break;
27         }
28     }
29     printf("%lld\n", sum);
30     for (int i = 1; i <= n; i++)
31         printf("%d%c", i, i == n ? '\n' : ' ');
32     for (int i = 1; i <= n; i++)
33         printf("%d%c", ans[i], i == n ? '\n' : ' ');
34 
35 }

posted @ 2019-10-14 13:05  祈梦生  阅读(106)  评论(0编辑  收藏