标签:memory should rip UI code span pre cccccc strong
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
You want to arrange n integers a1,?a2,?...,?an in some order in a row. Let‘s define the value of an arrangement as the sum of differences between all pairs of adjacent integers.
More formally, let‘s denote some arrangement as a sequence of integers x1,?x2,?...,?xn, where sequence x is a permutation of sequence a. The value of such an arrangement is (x1?-?x2)?+?(x2?-?x3)?+?...?+?(xn?-?1?-?xn).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence x that corresponds to an arrangement of the largest possible value.
Input
The first line of the input contains integer n (2?≤?n?≤?100). The second line contains n space-separated integers a1, a2, ..., an (|ai|?≤?1000).
Output
Print the required sequence x1,?x2,?...,?xn. Sequence x should be the lexicographically smallest permutation of a that corresponds to an arrangement of the largest possible value.
Sample Input
Input
5
100 -100 50 0 -50
Output
100 -50 0 50 -100
Hint
In the sample test case, the value of the output arrangement is (100?-?(?-?50))?+?((?-?50)?-?0)?+?(0?-?50)?+?(50?-?(?-?100))?=?200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence x1,?x2,?... ,?xp is lexicographically smaller than sequence y1,?y2,?... ,?yp if there exists an integer r(0?≤?r?<?p) such that x1?=?y1,?x2?=?y2,?... ,?xr?=?yr and xr?+?1?<?yr?+?1.
题意:
给定一个序列,求使得(x1?-?x2)?+?(x2?-?x3)?+?...?+?(xn?-?1?-?xn)最大且字典序最小的排列
化简式子,得到:x1-xn 即求出x1-xn最大的即可
然后按字典序最小的输出
代码:
#include<bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; int a[130]; int main() { int n; cin>>n; for(int i=0;i<n;i++){ cin>>a[i]; } int x,y,res=-INF; int xl,yl; for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ if(a[i]-a[j]>res){ x=a[i],y=a[j]; xl=i,yl=j; res=a[i]-a[j]; } if(a[j]-a[i]>res){ y=a[i],x=a[j]; xl=j,yl=i; res=a[j]-a[i]; } } } a[xl]=-INF,a[yl]=-INF; sort(a,a+n); cout<<x<<" "; for(int i=0;i<n;i++){ if(a[i]!=-INF){ cout<<a[i]<<" "; } } cout<<y<<endl; }
codefroces Round #201.a--Difference Row
标签:memory should rip UI code span pre cccccc strong
原文地址:http://www.cnblogs.com/liuzhanshan/p/6556863.html