原文鏈接:http://www.2cto.com/kf/201311/260148.html
一、康託展開:全排列到一個自然數的雙射
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#include<cstdio>
const
int
fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320}; ///階乘
int
KT( int
s[], int
n) {
int
i, j, cnt, sum; sum
= 0; for
(i = 0; i < n; ++i) {
cnt
= 0; for
(j = i + 1; j < n; ++j) if
(s[j] < s[i]) ++cnt; sum
+= cnt * fac[n - i - 1]; }
return
sum; }
int
main() {
int
a[] = {3, 5, 7, 4, 1, 2, 9, 6, 8}; printf ( "%d\n" ,
1 + KT(a, sizeof (a)
/ sizeof (*a)));
///1+98884
}
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#include<cstdio>
#include<cstring>
const
int
fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320}; ///階乘
bool
vis[10]; ///n爲ans大小,k爲全排列的編碼
void
invKT( int
ans[], int
n, int
k) {
int
i, j, t; memset (vis,
0, sizeof (vis));
--k;
for
(i = 0; i < n; ++i) {
t
= k / fac[n - i - 1]; for
(j = 1; j <= n; j++) if
(!vis[j]) {
if
(t == 0) break ;
--t;
}
ans[i]
= j, vis[j] = true ;
k
%= fac[n - i - 1]; ///餘數
}
}
int
main() {
int
a[10]; invKT(a,
5, 16); for
( int
i = 0; i < 5; ++i) printf ( "%d
" ,
a[i]); ///1
4 3 5 2 }
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