From e79e878400bd45c53c310e0e08d6233a472f90d0 Mon Sep 17 00:00:00 2001
From: Mohammad S Anwar <mohammad.anwar@yahoo.com>
Date: Sun, 8 Nov 2020 01:26:52 +0000
Subject: - Added solutions by Ulrich Rieke.

---
 challenge-085/ulrich-rieke/cpp/ch-2.cpp | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 challenge-085/ulrich-rieke/cpp/ch-2.cpp

(limited to 'challenge-085/ulrich-rieke/cpp/ch-2.cpp')

diff --git a/challenge-085/ulrich-rieke/cpp/ch-2.cpp b/challenge-085/ulrich-rieke/cpp/ch-2.cpp
new file mode 100644
index 0000000000..ed72f314bb
--- /dev/null
+++ b/challenge-085/ulrich-rieke/cpp/ch-2.cpp
@@ -0,0 +1,26 @@
+#include <iostream>
+#include <cmath>
+#include <cstdlib>
+
+int main( int argc, char * argv[] ) {
+  int n = std::atoi( argv[ 1 ] ) ;
+  int powerlimit = 0 ;
+  //the limit is the square root of n. We cycle through all numbers from 2 to
+  //the square root and finish cycling as soon as we find 2 integers that
+  //form n ; the functions from cmath require some type juggling
+  int lowerlimit = static_cast<int>( sqrt( static_cast<double>( n ) ) ) ;
+  for ( int i = 2 ; i < lowerlimit + 1 ; i++ ) {
+      double expo = 2.0 ;
+      int power = static_cast<int>(std::pow( static_cast<double>( i ) , expo )) ;
+      while ( power < n ) {
+    expo += 1.0 ;
+        power = static_cast<int>(std::pow( static_cast<double>( i ) , expo )) ;
+      }
+      if ( power == n ) {
+    powerlimit = 1 ;
+    break ;
+      }
+  }
+  std::cout << powerlimit << std::endl ;
+  return 0 ;
+}
-- 
cgit