Old Ieeextreme solutions

IEEExtreme11.0/Fill the Pixel.cpp

@@ -0,0 +1,112 @@
+#include<iostream>
+#include <cstring>
+using namespace std;
+int t,M,N;
+char screen1[1000][1000];
+char screen2[1000][1000];
+char screen3[1000][1000];
+// A recursive function to replace previous color 'prevC' at  '(x, y)'
+// and all surrounding pixels of (x, y) with new color 'newC' and
+void floodFillUtil1(int x, int y, int prevC, int newC)
+{
+    // Base cases
+    if (x < 0 || x >= M || y < 0 || y >= N)
+        return;
+    if (screen1[x][y] != prevC)
+        return;
+    if (screen1[x][y] == newC)
+        return;
+
+    // Replace the color at (x, y)
+    screen1[x][y] = newC;
+
+    // Recur for north, east, south and west
+    floodFillUtil1(x+1, y, prevC, newC);
+    floodFillUtil1(x-1, y, prevC, newC);
+    floodFillUtil1(x, y+1, prevC, newC);
+    floodFillUtil1(x, y-1, prevC, newC);
+}
+
+void floodFillUtil2(int x, int y, int prevC, int newC)
+{
+    // Base cases
+    if (x < 0 || x >= M || y < 0 || y >= N)
+        return;
+    if (screen2[x][y] != prevC)
+        return;
+    if (screen2[x][y] == newC)
+        return;
+
+    // Replace the color at (x, y)
+    screen2[x][y] = newC;
+
+    // Recur for north, east, south and west
+    floodFillUtil2(x + 1, y + 1, prevC, newC);
+    floodFillUtil2(x - 1, y - 1, prevC, newC);
+    floodFillUtil2(x - 1, y + 1, prevC, newC);
+    floodFillUtil2(x + 1, y - 1, prevC, newC);
+}
+
+
+void floodFillUtil3(int x, int y, int prevC, int newC)
+{
+    // Base cases
+    if (x < 0 || x >= M || y < 0 || y >= N)
+        return;
+    if (screen3[x][y] != prevC)
+        return;
+    if (screen3[x][y] == newC)
+        return;
+
+    // Replace the color at (x, y)
+    screen3[x][y] = newC;
+
+    // Recur for north, east, south and west
+    floodFillUtil3(x - 1, y - 1, prevC, newC);
+    floodFillUtil3(x - 1, y, prevC, newC);
+    floodFillUtil3(x - 1, y + 1, prevC, newC);
+    floodFillUtil3(x, y - 1, prevC, newC);
+    floodFillUtil3(x, y + 1, prevC, newC);
+    floodFillUtil3(x + 1, y - 1, prevC, newC);
+    floodFillUtil3(x + 1, y, prevC, newC);
+    floodFillUtil3(x + 1, y + 1, prevC, newC);
+}
+
+int main(){
+
+cin >> t;
+for (int i = 0; i <t; i++){
+    string str;
+    int c1 =0;
+    int c2 = 0;
+    int c3 = 0;
+    cin >> N >> M;
+    for (int j = 0; j<M; j++){
+        cin >> str;
+        for (int k = 0; k<N; k++){
+            screen1[j][k] = str.at(k);
+        }
+    }
+    memcpy (screen2, screen1, 1000*1000*sizeof(char));
+    memcpy (screen3, screen1, 1000*1000*sizeof(char));
+
+    for (int row =0; row < M; row++){
+        for (int col =0; col < N; col++){
+            if (screen1[row][col] == '1'){
+                floodFillUtil1(row, col, '1', '0');
+                c1 += 1;
+            }
+            if (screen2[row][col] == '1'){
+                floodFillUtil2(row, col, '1', '0');
+                c2 += 1;
+            }
+            if (screen3[row][col] == '1'){
+                floodFillUtil3(row, col, '1', '0');
+                c3 += 1;
+          }
+        }
+    }
+    cout << c1 << ' '<< c2 << ' ' << c3 << '\n';
+}
+
+}

IEEExtreme11.0/Math_challenge/math_challenge.cpp

@@ -0,0 +1,61 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+typedef unsigned long long ull;
+
+ull combo(ull n, ull k) { 
+    ull res = 1; 
+
+    // Since C(n, k) = C(n, n-k) 
+    if ( k > n - k ) 
+        k = n - k; 
+
+    // Calculate value of [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1] 
+    for (ull i = 0; i < k; ++i) 
+    { 
+        res *= (n - i); 
+        res /= (i + 1); 
+    } 
+
+    return res; 
+} 
+
+ull power(ull x, ull y) 
+{ 
+	const unsigned int M = 1000000007; 	
+    ull res = 1;      // Initialize result 
+
+    while (y > 0) 
+    { 
+        if (y%2 == 1) 
+            res = (res*x) % M; 
+        y = y/2; // y = y/2 
+        x = (x*x) % M;   
+    } 
+    return res; 
+}
+
+int main() {
+	int t;
+	ull a, b, c;
+
+	scanf("%d", &t);
+
+	while (t--) {
+		scanf("%llu%llu%llu", &a, &b, &c);
+
+
+		// ull res = 1;
+		ull comb = combo(b, c);
+		ull res = power(a, comb);
+
+		// for (ull i = 1; i <= comb; i++) 
+		// 	res = (res*a) % M; 
+
+		printf("%llu\n", res); 
+
+	}
+
+	return 0;
+}

IEEExtreme11.0/Math_challenge/math_challenge.py

@@ -0,0 +1,47 @@
+from math import floor
+
+def combo(n, k): 
+    # since C(n, k) = C(n, n - k) 
+    if(k > n - k): 
+        k = n - k 
+    # initialize result 
+    res = 1
+    # Calculate value of  
+    # [n * (n-1) *---* (n-k + 1)] / [k * (k-1) *----* 1] 
+    for i in range(k): 
+        res = res * (n - i) 
+        res = res / (i + 1) 
+    return res 
+
+
+def mpower(x, y) : 
+    res = 1     # Initialize result 
+    M = 1000000007 
+
+    while (y > 0) : 
+
+        # If y is odd, multiply 
+        # x with result 
+        if (y%2 ==  1) : 
+            res = (res * x) % M 
+
+        # y must be even now 
+        y = floor(y / 2) 
+        print(y)     # y = y/2 
+        x = (x * x) % M
+
+    return res 
+
+
+t = int(input())
+for i in range(t):
+	abc = input()
+	a_b_c = abc.split(" ")
+	a = int(a_b_c[0])
+	b = int(a_b_c[1])
+	c = int(a_b_c[2])
+
+	comb = combo(b, c)
+	res = mpower(a, floor(comb))
+	print(res)
+