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$x,y,z$
$x*w+y*d=p&& x+y+z=n&&xgeq 0&&ygeq 0&&zgeq 0$
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 "); 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 "); 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 "); 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 "); 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 ", Min); 53 dfs1(root, 0, ansi, 0); 54 for (int i = 1; i <= n; i++) 55 printf("%lld%c", ans[i], i == 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 ", 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$ans=sum_{i=1}^{n}max(p_{i},q_{i})&&ansleq k$
先固定一个序列$q$,为$1,2,3cdot cdot cdot n$,然后去构造序列$p$。
序列$p$初始也为$1,2,3cdot cdot cdot n$,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 "); 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 ", sum); 30 for (int i = 1; i <= n; i++) 31 printf("%d%c", i, i == n ? ' ' : ' '); 32 for (int i = 1; i <= n; i++) 33 printf("%d%c", ans[i], i == n ? ' ' : ' '); 34 35 }