UVA 11925 Generating Permutations（生成排列）（构造）

题意：将序列1,2,3,……,n，用不超过2n^2次操作，通过下列操作变成给定序列。(1<=n<=300)

1、交换前两个元素

2、将第一个元素移到最后

分析：因为将序列变成升序更容易操作，所以倒着输出解，进行如下操作：

1、交换前两个元素

2、将最后一个元素移到第一个(当a[0] < a[1] || (a[0] == n && a[1] == 1))

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 300 + 10;
const int MAXT = 10000 + 10;
using namespace std;
vector<int> a, ans;
int n;
bool judge(){
    for(int i = 0; i < n; ++i){
        if(a[i] != i + 1) return false;
    }
    return true;
}
int main(){
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        a.clear();
        ans.clear();
        for(int i = 0; i < n; ++i){
            int x;
            scanf("%d", &x);
            a.push_back(x);
        }
        while(1){
            if(judge()) break;
            if(a[0] < a[1] || (a[0] == n && a[1] == 1)){
                ans.push_back(2);
                a.insert(a.begin(), a[n - 1]);
                a.erase(a.end() - 1);
            }
            else{
                ans.push_back(1);
                swap(a[0], a[1]);
            }
        }
        int len = ans.size();
        for(int i = len - 1; i >= 0; --i) printf("%d", ans[i]);
        printf("\n");
    }
    return 0;
}