CodeForces 686B - Little Robber Girl's Zoo

题意：将一个n（1 <= n <= 100）个元素的序列排成非递减序列，每次操作可以指定区间[ L，R ]（区间内元素个数为偶数），将区间内第一项与第二项交换，第三项与第四项交换，第五项与第六项……在2W次内完成排序，输出每次操作。

瞎搞即可，不断检查相邻元素是否满足 前者>=后者，不满足即交换，直到序列满足条件位置（n最大为100绝对不会超过2W步）

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << " "
#define prln(x) cout << #x << " : " << x << endl
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {-1, 1, 0, 0};
const ll MOD = 1e9 + 7;
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;

int n, a[MAXN];

int main(){
while(scanf("%d", &n) == 1){
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
bool flag = true;
while(flag){
flag = false;
for(int i = 1; i < n; ++i)
if(a[i] > a[i + 1]){
flag = true;
swap(a[i], a[i + 1]);
printf("%d %d\n", i, i + 1);
}
}
}
return 0;
}