#define L 5
int N[L] = {
1,
2,
3,
4,
5};
void print_permutation(
int n,
int *a,
int cur) {
if ( cur ==
n ) {
for (
int i =
0; i < n; i++
) {
printf("%d ", a[i]);
}
printf("\n");
}
else {
for (
int i =
0; i < n; i++
) {
int has =
0;
for (
int j =
0; j < cur; j++
) {
if ( a[j] == N[i] ) has =
1;
}
if (!
has ) {
a[cur] =
N[i];
print_permutation(n,a,cur+
1);
}
}
}
}
int main() {
int a[L] = {
0};
print_permutation(L, a, 0);
return 0;
}
还可以直接使用C++提供的STL （原来还可以这样啊摔！）
int main () {
int n, p[
10];
scanf("%d", &
n);
for (
int i =
0; i < n; i++) scanf(
"%d", &
p[i]);
sort(p, p+
n);
do {
for (
int i =
0; i < n; i++ ) printf(
"%d", p[i]);
printf("\n");
} while(next_permutation(p, p+
n));
system("PAUSE");
return 0;
}
转载于:https://www.cnblogs.com/tsubasa/archive/2012/12/06/2805426.html
