SPOJ_FIBOSUM

SPOJ_FIBOSUM

@@ -0,0 +1,76 @@
+//Fibonacci Sum
+//https://www.spoj.com/problems/FIBOSUM/
+//This question wants the user to calculate Fibonacci with in a range. A naive approach will be to simply use the
+//relation F(N) = F(N - 1) + F(N - 2) to find the nth fibonacci and the itterate over the range. A better approach will be to derrive
+//a recurence relation of the fibonacci series. Refer this link for mathematics behind it https://www.youtube.com/watch?v=WT_TGxQrV1k&t=214s
+//Even now itterating to find the sum for fibbonaci will give us TLE because of large constraints.
+//So there's an observation to get TLE fixed. 1, 1, 2, 3, 5, 8, 13, 21, 34. The observation is Sum till Nth fibonacci is F(n+2) - 1
+//This observation is true throughout. Now calculating sum within range from n to m will be sum till m minus sum till n
+
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long int lli;
+#define mat(x, y, name) vector< vector<lli> > name (x, vector<lli>(y));
+
+lli MOD = 1000000007;
+
+vector< vector<lli> > matMul(vector< vector<lli> > A, vector< vector<lli> > B)
+{
+    vector< vector<lli> > C(A.size(), vector<lli>(B[0].size()));
+    for (int i = 0; i < A.size(); i++)
+    {
+        for (int j = 0; j < B[0].size(); j++)
+        {
+            C[i][j] = 0;
+            for (int k = 0; k < B.size(); k++)
+                C[i][j] = (C[i][j] + ((A[i][k] * B[k][j]) % MOD)) % MOD;
+        }
+    }
+    return C;
+}
+
+vector< vector<lli> > matPow(vector< vector<lli> > A, int p)
+{
+    if (p == 1)
+        return A;
+    if (p&1)
+        return matMul(A, matPow(A, p-1));
+    else
+    {
+        vector< vector<lli> > C = matPow(A, p/2);
+        return matMul(C, C);
+    }
+}
+
+int main()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    mat(2, 1, F);
+    mat(2, 2, T);
+    T[0][0] = 0;
+    T[0][1] = 1;
+    T[1][0] = 1;
+    T[1][1] = 1;
+    F[0][0] = 1;
+    F[1][0] = 1;
+    int t;
+    cin >> t;
+    while(t--)
+    {
+        lli n, m;
+        cin >> n >> m;
+        lli minSum = (n == 0) ? 0 : ((matMul(matPow(T, n), F)[0][0]) - 1) % MOD;
+        lli maxSum = (m == 0) ? 0 : ((matMul(matPow(T, m+1), F)[0][0]) - 1) % MOD;
+        lli ans = (maxSum - minSum) % MOD;
+        if (ans < 0)
+        {
+            ans += MOD;
+            ans = ans % MOD;
+        }
+        cout << ans << endl;
+    }
+    return 0;
+}