MemSQL problem A & B

Author: Shuo Xiang

Codeforces/335A.py

@@ -0,0 +1,35 @@
+import sys
+
+
+s = raw_input()
+n = input()
+
+d = [0] * 40
+for i in s:
+	d[ord(i) - 97] += 1
+
+sn = len(filter(lambda x: x > 0, d))
+		
+if n < sn:
+	print -1
+else:
+	for ans in xrange(1, len(s) + 1):
+		tmp = 0
+		for i in xrange(40):
+			tmp += (d[i] + ans - 1) / ans
+		if tmp <= n:
+			cnt = 0
+			print ans
+			for i in xrange(40):
+				for j in xrange ((d[i] + ans - 1) / ans):
+					cnt += 1
+					sys.stdout.write(chr(i + 97))
+			while cnt < n:
+				sys.stdout.write('a')
+				cnt += 1
+			break
+
+
+
+			
+		

Codeforces/335B.cpp

@@ -0,0 +1,84 @@
+#include <cstdio>
+#include <algorithm>
+#include <iostream>
+#include <string>
+
+using namespace std;
+
+const int N = 50011;
+
+int d[2611][2611];
+int pa[2611][2611];
+int cnt[26];
+string s, ans;
+
+void output (int x, int y) {
+	if (x > y) return;
+	if (x == y) { ans += s[x-1]; return;}
+	
+	if (!pa[x][y]) {
+		ans += s[x-1];
+		output (x + 1, y - 1);
+		ans += s[y-1];
+	}
+	else if (pa[x][y] == 1)
+		output (x + 1, y);
+	else
+		output (x, y - 1);
+}
+
+int main () {
+	int i, j, k;
+	cin >> s;
+	int sn = s.size ();
+	for (i = 0; i < sn; cnt[s[i++] - 'a'] ++);
+
+	for (i = 0; i < 26; ++i) {
+		if (cnt[i] >= 100) {
+			for (j = 0; j < 100; ++j)
+				printf ("%c", i + 'a');
+			printf ("\n");
+			return 0;
+		}
+	}
+
+	int x, y, m = -1;
+			
+	for (k = 1; k <= sn; ++k)
+		for (i = 1; i + k - 1 <= sn; ++i) {
+			j = i + k - 1;
+			if (k == 1)
+				d[i][k] = 1, pa[i][j] = 1;
+			else {			
+				if (d[i+1][k-1] >= d[i][k-1])
+					pa[i][j] = 1;
+				else
+					pa[i][j] = -1;
+				int ma = max (d[i+1][k-1], d[i][k-1]);
+				if (s[i-1]==s[j-1] && 2 + d[i+1][k-2] >= ma) {
+					ma = 2 + d[i+1][k-2];
+					pa[i][j] = 0;
+				}
+				d[i][k] = ma;
+				if (d[i][k] == 100) {
+					output (i, j);
+					cout << ans << endl;
+					return 0;
+				}
+			}
+			if (d[i][k] > m) 
+				m = d[i][k], x = i, y = j;				
+		}
+	output (x, y);
+	sn = ans.size();
+	if (sn > 100) {
+		for (i = 0; i < 50; ++i)
+			cout << ans[i];
+		for (i = sn - 50; i <= sn - 1; ++i)
+			cout << ans[i];
+		cout << endl;
+	}
+	else
+		cout << ans << endl;
+	return 0;
+}

Codeforces/Abbyy/331A2.cpp

@@ -0,0 +1,61 @@
+#include <cstdio>
+#include <algorithm>
+#include <map>
+#include <cmath>
+#include <cstring>
+
+
+using namespace std;
+
+const int N = 300011;
+map<int,int> ord, rev;
+long long possum[N];
+int cntneg[N];
+
+int a[N], n;
+
+long long f(int x, int y) {
+	return possum[y-1]-possum[x] + a[x] + a[y];
+}
+
+int g(int x, int y) {
+	return cntneg[y-1]-cntneg[x];
+}
+
+int main () {
+	int i, j, k;
+	scanf ("%d", &n);
+	memset(cntneg, 0, sizeof(cntneg));
+	memset(possum, 0, sizeof(possum));
+	
+	for (i = 1; i <= n; ++i) {
+		scanf ("%d", &a[i]);
+		possum[i] = possum[i-1] + max(a[i], 0);
+		cntneg[i] = cntneg[i-1] + (a[i] < 0 ? 1 : 0);
+		if (ord.find(a[i]) != ord.end()) continue;
+		else ord[a[i]] = i;
+	}
+	for (i = n; i >= 1; --i) {
+		if (rev.find(a[i]) != rev.end()) continue;
+		else rev[a[i]] = i;
+	}
+	int bestx, besty;
+	long long best = -1000000000000000;
+	for (i = 1; i <= n; ++i) {
+		int x = ord[a[i]], y = rev[a[i]];
+		if (x != i || ord[a[i]] == rev[a[i]]) continue;
+		long long t;
+		if ((t = f(x, y)) > best) {
+			best = t;
+			bestx = x;
+			besty = y;
+		}
+		
+	}
+	printf ("%I64d %d\n", best, g(bestx, besty) + bestx - 1 + n - besty);
+	for (i = 1; i < bestx; printf("%d ", i++));
+	for (i = bestx + 1; i < besty; ++i)
+		if (a[i] < 0) printf ("%d ", i);
+	for (i = besty+1; i <= n; printf("%d ", i++));
+	return 0;
+}

Codeforces/Div II/CF327C_191.scala

@@ -0,0 +1,25 @@
+object CF327C_191 extends App {
+  val s = Console.readLine
+  val n : Long = s.size.toLong
+  val k : Long = Console.readLine.toLong
+  val md = 1000000007L
+
+  def p2(y : Long, x : Long) : Long = {
+    if (x == 0) 1 else {
+      val t = p2(y, x/2)
+      if (x % 2 == 0) (t * t) % md else (((y * t) % md) * t) % md
+    }
+  }
+
+  def inv(x : Long) : Long = {
+    p2(x, md - 2)
+  }
+
+  val res = (0 until n.toInt).map{
+    x => if (s(x) == '0' || s(x) == '5') p2(2, x) else 0
+  }.reduce((x, y) => (x + y) % md)
+
+  val outer = (p2(2, k * n) - 1 + md) % md * inv((p2(2, n) - 1 + md)% md) % md
+
+  println (res * outer % md)
+}