Kanto expansion (used for Full Permutation and Integer Conversion)

Kanto expansion formula

X = An * (n-1 )! + An-1 * (n-2 )! +... + AI * (I-1 )! +... + A2 * 1! + A1 * 0! Where, A is an integer and 0 <= AI <I (1 <= I <= N ). This is Kanto.

Expand instance {1, 2, 3, 4 ,..., n} indicates 1, 2, 3 ,..., for example, {1, 2, 3} is arranged in ascending order of N. 123 132 213 231 312 321.

The number 1, 2, 4, 5, and 6 correspond to a given number.

The relationships between them can be found in Kanto.

If you want to know the number of the largest numbers in 321 is {1, 2, 3}, you can consider this as follows:

The first digit is 3. When the first digit is less than 3, the number is smaller than 321, for example, 123 and 213. The number smaller than 3 is 1 and 2. So there are 2*2! . Let's look at the number smaller than the second digit 2: only one number smaller than 2 is 1, so there is 1*1! = 1 so the number of {1, 2, 3} arrays smaller than 321 is 2*2! + 1*1! = 5. So 321 is a large number of 6th. 2*2! + 1*1! + 1*0! Is to expand the Kanto.

For another example, 1324 is the largest number in the {1, 2, 3, 4} number. The first is the number where 1 is smaller than 1. The number is 0 0*3! The second digit is 3 less than 3. The numbers are 1 and 2, but 1 is already in the first place, so there is only one digit 2 1*2! . The third digit is 2 and the number smaller than 2 is 1, but 1 is in the first place, so there are 0 numbers 0*1! So there are 0*3 smaller than 1324! + 1*2! + 0*1! = 2, 1324 is the third large number. Code

const int PermSize = 12;long long factory[PermSize] = { 0, 1, 2, 6, 24, 120,720, 5040, 40320, 362880, 3628800,39916800 };long long Cantor(string buf) {int i, j, counted;long long result = 0;for (i = 0; i < PermSize; ++i) {counted = 0;for(j = i + 1; j < PermSize; ++j)if(buf[i] > buf[j])++counted;result = result + counted *factory[PermSize - i - 1];}return result;}

Source: Baidu encyclopedia