辗转相除，计算m * x mod n = 1，已知m < n, 并且m与n互质，求x的值。

mod.js

function count(m, n){

  if (n < m) return;

  var mod = Math.floor( n / m)
  var rem = n % m
  var number = 1

  var paths = []

  paths.push([mod, rem, number])

  mod += 1
  rem -= m

  paths.push([mod, rem, number])

  if (rem == -1) return paths

  add(paths)

  var closes = findClose(paths)

  while(closes.distance !== -1) {
    add(paths, closes.route)
    closes = findClose(paths)
  }

  add(paths, closes.route)

  return paths
}

function add(paths, route){
  route = route || [0, 0]
  var exp1 = paths[route[0]]
  var exp2 = paths[route[1]]

  var mod = exp1[0] + exp2[0]
  var rem = exp1[1] + exp2[1]
  var number = exp1[2] + exp2[2]

  paths.push([mod, rem, number])
}

function findClose(paths){

  var minPositive = 0
  var maxNegative = 0
  var ret = [0, 0]

  paths.forEach(function(num, i){

    var rem = num[1]

    if (rem > 0) {

      // 如果最小正数为0，那么初始化minPositive
      if (minPositive === 0) {
        minPositive = rem
        ret[0] = i
      }

      // 寻找更小的
      if (minPositive > rem) {
        minPositive = rem
        ret[0] = i
      }

    }

    if (rem < 0) {

      // 如果最小正数为0，那么初始化minPositive
      if (maxNegative === 0) {
        maxNegative = rem
        ret[1] = i
      }

      // 寻找更小的
      if (maxNegative < rem) {
        maxNegative = rem
        ret[1] = i
      }

    }

  })

  return { route: ret, distance: minPositive + maxNegative }

}

console.log(count(3, 10000))