**Due: Oct 26 at 11:55 pm**

Not really a LeetCode problem, since Leetcode doesn't have this problem.

Instead we use Geeks for Geeks Practice

Your goal is to write a divide and conquer solution for Modular Exponentiation for large numbers

Read about Modular exponentiation here: Modular exponentiation

So we want to be able to compute *b^e % m* where *b* is base, *e* is the exponent, and *m* is the modulus.

Modular exponentiation is especially useful in the field of public-key cryptography. The problem is that for strong cryptography we need *b* to be at least 256 bits long! Moreover, we have to take this large number to an exponent, which can get REALLY big!

Example: (234^432) mod 311 = 140

This cannot possibily be computed directly with (32 bit) int's or even long's in C++. Also - we need to use integers, as using doubles will not work because of round-off error.

Fortunately we can use the following modular property to break a given modular exponentiation into smaller problems.

cmodm= (a⋅b) modmcmodm= [(amodm) ⋅ (bmodm)] modm

You goal is to write a ** divide and conquer algorithm** to do modular exponentiation. The idea is to use the property above to divide the problem into smaller probems that are computible with int's without overflowing. Try to build a solution that minimizes the magnitude of the intermediate results as much as possible.

The C++ **pow()** function in math.c works on doubles, which you should not use. However, the following integer **intPow()** function can be used instead:

intintPow(intbase,intexp) {intresult = 1;while(exp) {if(exp & 1) result *= base; exp /= 2; base *= base; }returnresult; }

base | exponent | modulus | b^e % m |

34523 | 23 | 713 | 391 |

43565 | 335 | 271 | 114 |

43311 | 3661 | 51 | 48 |

544654 | 325 | 5511 | 913 |

12343 | 32123 | 211 | 83 |

4354 | 211 | 23 | ? |

24411 | 331 | 533 | ? |

43554 | 511 | 411 | ? |

54324 | 35542 | 255 | ? |

34563 | 543255 | 5331 | ? |

- Complete working program code
- Runs for all cases above
- A tight runtime analysis

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