Added POLYMUL from SPOJ

Author: rishabhdeepsingh

PolyMul/question.txt

@@ -0,0 +1,31 @@
+Sam and Dean fight the supernatural creatures to protect the humans. Now they have come across a creature called the Vedala, which are always found in pairs. Each vedala can be represented by a polynomial. Both the Vedalas need to be killed at once or else they can't be killed. They can be killed only by using the product of the two polynomials representing the two Vedala. Help Sam and Dean find the product of the polynomials fast, so they can do thier work. 
+
+Input
+
+First line contains an integer T (≤ 10), number of test cases.
+Each test case will have n (n ≤ 10000), maximum degree of the polynomials on the first line. 
+
+Next two lines will have n+1 space separated integers each representing the coeffiecients of first and second polynomials respectively.
+
+All input coefficients values are <=1000.
+
+Output
+For each test case ouput a line of 2n space seperated integers indicating coefficients of the polynomial created after multiplication.
+
+
+Example 
+Input:
+2
+2
+1 2 3
+3 2 1
+2
+1 0 1
+2 1 0
+
+Output:
+3 8 14 8 3
+2 1 2 1 0
+
+
+

PolyMul/solution.cpp

@@ -0,0 +1,86 @@
+#include<bits/stdc++.h>
+
+using namespace std;
+
+typedef long long int LL;
+typedef vector<LL> VLL;
+typedef std::complex<double> CD;
+
+#define PB              push_back
+#define F               first
+#define s               second
+#define SZ(x)           x.size()
+#define ALL(a)          std::begin(a), std::end(a)
+#define IN_REP          int _t; cin >> _t ; while(_t--)
+#define IOS             ios::sync_with_stdio(false);cin.tie(NULL)
+#define FOR(i, a, b)    for(int i=(a);i<(b);i++)
+#define REP(i, n)       FOR(i,0,n)
+const double PI     =   acos(-1);
+typedef std::valarray<CD> CArray;
+
+// Cooley–Tukey FFT (in-place)
+void fft(CArray &x) {
+    // const size_t N = x.size();
+    int N = SZ(x);
+    if (N <= 1) return;
+
+    // divide
+    CArray even = x[std::slice(0, N / 2, 2)];
+    CArray odd = x[std::slice(1, N / 2, 2)];
+
+    // conquer
+    fft(even);
+    fft(odd);
+
+    // combine
+    // for (size_t k = 0; k < N/2; ++k){
+    for (int k = 0; k < N / 2; ++k) {
+        CD t = std::polar(1.0, -2 * PI * k / N) * odd[k];
+        x[k] = even[k] + t;
+        x[k + N / 2] = even[k] - t;
+    }
+}
+
+// inverse fft (in-place)
+void ifft(CArray &x) {
+    // conjugate the complex numbers
+    x = x.apply(std::conj);
+
+    // forward fft
+    fft(x);
+
+    // conjugate the complex numbers again
+    x = x.apply(std::conj);
+
+    // scale the numbers
+    x /= x.size();
+}
+
+int main() {
+    IOS;
+    IN_REP {
+        int n;
+        cin >> n;
+        n++; // Given in question
+        int size = 2 * (1 << int(ceil(log2(n))));
+
+        CArray x(size), y(size);
+        FOR(i, size - n, size) cin >> x[i];
+        fft(x);
+        FOR(i, size - n, size) cin >> y[i];
+        fft(y);
+
+        CArray res(size);
+        res = x * y;
+        ifft(res);
+        VLL ans;
+        REP(i, size - 1) {
+            ans.PB(round(res[i].real()));
+        }
+        for (int i = size - 1 - (2 * n - 1); i < size - 1; i++) {
+            cout << ans[i] << " ";
+        }
+        cout << endl;
+    }
+    return 0;
+}