Number Painting

K.py

@@ -0,0 +1,35 @@
+def gcd(a,b):
+	if(a==0):
+		return b
+	else:
+		return gcd(b%a,a)
+
+from math import *
+k,n=map(int,input().split())
+if(k==n-1):
+	print(0)
+elif(k==n):
+	print('1/'+str(factorial(n)))
+else:
+	fact=[1,1,2]
+	for i in range(3,max(n-k,k)+1):
+		fact.append(fact[-1]*i)
+
+	if((n-k)%2==0):
+		num=1
+	else:
+		num=-1
+	prod=1
+	for i in range(n-k,0,-1):
+		prod*=i
+		if(i%2==1):
+			num+=prod
+		else:
+			num-=prod
+
+	den=fact[k]*fact[n-k]
+
+	g=gcd(num,den)
+	num=num//g
+	den=den//g
+	print(str(num)+"/"+str(den))

Solutions/101503 - 2007-2008 ACM-ICPC, NEERC, Southern Subregional Contest/J.cpp

@@ -0,0 +1,68 @@
+
+#include <bits/stdc++.h>
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+#include <ext/pb_ds/detail/standard_policies.hpp>
+using namespace std;
+using namespace __gnu_pbds;
+
+typedef vector <long long> vi;
+typedef pair <long long, long long> pii;
+#define pb push_back   
+#define all(c) c.begin(), c.end()
+#define For(i, a, b) for (long long i = a; i < b; ++i)
+#define Forr(i, a, b) for (long long i = a; i > b; --i)
+#define um unordered_map
+#define F first
+#define S second
+#define int long long
+#define inf LLONG_MAX
+#define endl "\n"
+#define min(a,b) (a < b ? a : b)
+#define max(a,b) (a > b ? a : b)
+#pragma comment(linker, "/stack:200000000")
+#pragma GCC optimize ("O3")
+#pragma GCC optimize ("unroll-loops")
+#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
+#define _CRT_SECURE_NO_WARNINGS
+
+int modulo = 1e9 + 7;
+
+int fpow(int a, int n) {
+        int ans = 1;
+        while (n) {
+                if (n&1) ans = (ans*a)%modulo;
+                a = (a*a)%modulo; n = n >> 1;
+        }
+        return ans;
+}
+
+signed main() {
+        ios_base::sync_with_stdio(false);
+        cin.tie(NULL);
+        cout.tie(NULL);
+        // #ifndef ONLINE_JUDGE
+        // freopen("input.txt", "r", stdin);
+        // freopen("output.txt", "w", stdout);
+        // #endif
+        int n; cin >> n;
+       	int ans[n];
+       	int y = 0;
+       	For(i,0,n) {
+       		int num = i+1;
+       		int val = 1;
+       		for (int j = 1; j <= sqrt(num); ++j) {
+       			if (num%j == 0 and j != num) {
+       				val = max(val, ans[j-1]+1);
+       				int x = num/j;
+       				if (x != num)
+       					val = max(val, ans[x-1]+1);
+       			}
+       		}
+       		ans[i] = val;
+       		y = max(y, ans[i]);
+       	}
+       	cout << y << endl;
+       	printarr(ans,n);
+        return 0;
+}