Added SPOJ problems

Author: kushagrasaksena

SPOJ/Arranging Amplifiers.cpp

@@ -0,0 +1,81 @@
+/* SPOJ */
+/* Title - Arranging Amplifiers */
+/* Created By - Kushagra Saxena */
+/* Date - 04/10/2020 */
+
+//Problem link - https://www.spoj.com/problems/ARRANGE/
+
+
+#include<bits/stdc++.h>
+using namespace std;
+#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
+#define rep(i,a,b) for(ll i=a; i<b; i++)
+#define per(i,a,b) for(ll i=a; i>=b; i--)
+#define ll long long int
+#define ld long double
+#define vi vector<ll>
+#define vii vector <pair<ll,ll> >
+#define pb push_back
+#define mkp make_pair
+#define ff first
+#define ss second
+#define MOD 1000000007
+#define umap unordered_map<ll, ll>
+#define map map<ll, ll>
+#define autoit(x,it) for(auto it = x.begin(); it != x.end(); it++)
+#define mems(a, i) memset(a, i, sizeof(a))
+#define endl '\n'
+#define all(v) v.begin(),v.end()
+#define gcd(a,b) __gcd((a),(b))
+#define lcm(a,b) ((a)*(b))/gcd((a),(b))
+#define deba(a) cout << #a << " " << a << endl;
+int main()
+{
+	fast;
+	int t;
+	cin >> t;
+	while (t--)
+	{
+		ll n;
+		cin >> n;
+		vi a1, a2, a3;
+		rep(i, 0, n)
+		{
+			ll x;
+			cin >> x;
+			if (x == 1)
+				a1.pb(1);
+			else if (x == 2 or x == 3)
+				a2.pb(x);
+			else
+				a3.pb(x);
+		}
+		vi ans;
+		if (a1.size() > 0)
+		{
+			rep(i, 0, a1.size())
+			ans.pb(a1[i]);
+		}
+		ll fl = 0;
+		if (a3.size() > 0)
+		{
+			fl = 1;
+			sort(all(a3), greater<>());
+			rep(i, 0, a3.size())
+			ans.pb(a3[i]);
+		}
+		if (a2.size() > 0)
+		{
+			if (fl == 1)
+				sort(all(a2), greater<>());
+			else
+				sort(all(a2));
+			rep(i, 0, a2.size())
+			ans.pb(a2[i]);
+		}
+		rep(i, 0, ans.size())
+		cout << ans[i] << " ";
+		cout << endl;
+	}
+	return 0;
+}

SPOJ/The last digit.cpp

@@ -0,0 +1,61 @@
+/* SPOJ */
+/* Title - Arranging Amplifiers */
+/* Created By - Kushagra Saxena */
+/* Date - 04/10/2020 */
+
+//Problem link - https://www.spoj.com/problems/LASTDIG/
+
+
+// Nestor was doing the work of his math class about three days but he is tired of make operations a lot and he 
+// should deliver his task tomorrow. His math’s teacher gives him two numbers a and b. The problem consist of finding 
+// the last digit of the potency of base a and index b. Help Nestor with his problem. You are given two integer numbers:
+//  the base a (0 <= a <= 20) and the index b (0 <= b <= 2,147,483,000), a and b both are not 0.
+//  You have to find the last digit of ab.
+
+// Input
+// The first line of input contains an integer t, the number of test cases (t <= 30). t test cases follow. 
+// For each test case will appear a and b separated by space.
+
+// Output
+// For each test case output an integer per line representing the result
+
+// Sample Input
+// Input:
+// 2
+// 3 10
+// 6 2
+
+// Output:
+// 9
+// 6
+
+
+#include<bits/stdc++.h>
+using namespace std;
+#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
+#define ll long long int
+ll modularExpo(ll a, ll n, ll MOD)
+{
+	ll res = 1;
+	while (n > 0)
+	{
+		if (n % 2 == 0)
+			a = (a * a) % MOD, n /= 2;
+		else
+			res = (res * a) % MOD , n--;
+	}
+	return res;
+}
+int main()
+{
+	fast;
+	int t;
+	scanf("%d", &t);
+	while (t--)
+	{
+		ll a, b;
+		scanf("%lld%lld", &a, &b);
+		printf("%lld\n", modularExpo(a, b, 10));
+	}
+	return 0;
+}