Project Euler Problem 24
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 using namespace std;
5
6 int main(){
7 int factorial[9];
8 int sign[9];
9 int mul = 1;
10 int sum = 999999;
11 int tmp;
12 vector<int> out;
13
14 for(int i=0; i<10; i++){
15 out.push_back(i);
16 }
17
18 for(int i=1; i<10; i++){
19 mul = mul*i;
20 factorial[9-i] = mul;
21 }
22
23 for(int i=0; i<9; i++){
24 sign[i] = sum/factorial[i];
25 sum -= sign[i]*factorial[i];
26 }
27
28 for(int i =0; i<9; i++){
29 cout << i << '\t' << sign[i] << '\t' << factorial[i] << endl;
30 }
31 cout << endl;
32
33 for(int i=0; i< 9; i++)
34 {
35 tmp = out[i];
36 out[i] = out[sign[i]+i];
37 out[sign[i]+i] = tmp;
38 sort(out.begin()+i+1,out.end());
39 for(int j=0; j<10; j++){
40 cout << out[j];
41 }
42 cout << endl;
43 }
44
45 cin.get();
46 return 0;
47 }
2 #include <vector>
3 #include <algorithm>
4 using namespace std;
5
6 int main(){
7 int factorial[9];
8 int sign[9];
9 int mul = 1;
10 int sum = 999999;
11 int tmp;
12 vector<int> out;
13
14 for(int i=0; i<10; i++){
15 out.push_back(i);
16 }
17
18 for(int i=1; i<10; i++){
19 mul = mul*i;
20 factorial[9-i] = mul;
21 }
22
23 for(int i=0; i<9; i++){
24 sign[i] = sum/factorial[i];
25 sum -= sign[i]*factorial[i];
26 }
27
28 for(int i =0; i<9; i++){
29 cout << i << '\t' << sign[i] << '\t' << factorial[i] << endl;
30 }
31 cout << endl;
32
33 for(int i=0; i< 9; i++)
34 {
35 tmp = out[i];
36 out[i] = out[sign[i]+i];
37 out[sign[i]+i] = tmp;
38 sort(out.begin()+i+1,out.end());
39 for(int j=0; j<10; j++){
40 cout << out[j];
41 }
42 cout << endl;
43 }
44
45 cin.get();
46 return 0;
47 }
http://www.cnblogs.com/xianglan/archive/2011/03/07/1976431.html
