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CodeForces 686B - Little Robber Girl's Zoo(暴力)

时间:2016-10-25 13:43:25      阅读:198      评论:0      收藏:0      [点我收藏+]

标签:ring   int   cst   暴力   iterator   stream   cout   little   lis   

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;  
}  

 

CodeForces 686B - Little Robber Girl's Zoo(暴力)

标签:ring   int   cst   暴力   iterator   stream   cout   little   lis   

原文地址:http://www.cnblogs.com/tyty-TianTengtt/p/5996032.html

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