What I've done today is Self powers in Rust and Summary Ranges in JavaScript.

Math

Problem

Consecutive prime sum

Problem 50

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

Solution

Brute force.

Implementation

extern crate primal;

use primal::Sieve;

fn main() {
    const LIMIT: usize = 1_000_000;
    let sieve: Sieve = Sieve::new(LIMIT);
    let bound = sieve.prime_pi(LIMIT);
    let mut ans = 0;
    let mut cnt = 0;
    for i in 1..bound {
        let mut sum = 0;
        let mut offset = 0;
        if i + cnt > bound {
            break
        }
        while sum < LIMIT {
            if offset > cnt && sieve.is_prime(sum) {
                ans = sum;
                cnt = offset;
            }
            sum += sieve.nth_prime(i + offset);
            offset += 1;
        }
    }
    println!("Answer is {}", ans);
}

Algorithm

Problem

228. Summary Ranges

Given a sorted integer array without duplicates, return the summary of its ranges.

Example 1:

Input:  [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.

Example 2:

Input:  [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.

Solution

Nothing to say.

Implementation

/**
 * @param {number[]} nums
 * @return {string[]}
 */
var summaryRanges = function(nums) {
  if (nums.length === 0) return [];
  let ans = [];
  let begin = nums[0];
  let cur = nums[0];
  for (let i = 1; i < nums.length; ++i) {
    if (nums[i] - cur === 1) cur += 1;
    else {
      if (cur === begin) ans.push(cur+"");
      else ans.push(begin+"->"+cur);
      begin = cur = nums[i];
    }
  }
  if (cur === begin) ans.push(cur+"");
  else ans.push(begin+"->"+cur);
  return ans;
};