updated

Author: sameer2800

1189-sum_of_factorials.cpp

@@ -0,0 +1,77 @@
+#include <bits/stdc++.h>
+#define ll long long
+#define MAX 100100
+
+using namespace std;
+
+ll arr[22];
+ll facto[22];
+vector <pair <ll, ll> > v;
+
+void fact()
+{
+	facto[0] = 1;
+
+	for (int i = 1; i < 21; i++)
+		facto[i] = i * facto[i - 1];
+}
+
+ll yes ;
+ll indx;
+
+void recursive(ll num,  ll pos)
+{
+//	cout << num << endl;
+	if (num == 0 )
+	{
+		yes = 1;
+		return;
+	}
+
+	if (pos == -1)
+		return;
+
+	if (num >= facto[pos]) {
+		arr[indx] = pos;
+		indx++;
+		recursive(num - facto[pos], pos - 1 );
+		return;
+	}
+
+	recursive(num, pos - 1);
+}
+
+
+
+int main()
+{
+#ifndef ONLINE_JUDGE
+	freopen("/home/sameer/Documents/sameer/input.sam", "r", stdin);
+#endif
+	ios_base::sync_with_stdio(false);
+	ll i, j, k, n, m, t, cont, a, b;
+	cin >> t;
+	ll cases = t;
+	fact();
+
+	while (t--) {
+		cin >> n ;
+		yes = 0;
+		indx = 0;
+		memset(arr, 0, sizeof arr);
+		recursive(n,  20);
+		cout << "Case " << cases - t << ": "  ;
+
+		if (yes == 1) {
+			for (i = indx - 1; i >  0; i--)
+				cout << arr[i] << "!+";
+
+			cout << arr[0] << "!\n";
+
+		} else {
+			cout << "impossible\n";
+		}
+	}
+
+	return 0;
+}

README.md

@@ -470,6 +470,16 @@ and continue the loop ,till we find n as odd number.
 	one by one.
 
 
+##Solution 1189 :
+
+	As the time limit is too low 0.5 sec,if we do bitmasking , it will 
+	be timeout .
+	if we observe carefully ,
+	fact[n] > fact[n-1]  + fact[n-2] + ..... fact[1] + fact[0].
+	which means , if , k > fact[n] , and if k is a valid solution , then
+	k must contain fact[n] .
+	so just we need a single recursive call instead of double call inside 
+	the recursive function.
 
 ##Solution 1233 :