枚举生成1-n的排列

#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;
}