Codeforces Round #182 (Div. 2) C

有點意思的題目。。。

解題的關鍵在於， 假設這個數組中負數的個數為N，非負數的個數為M ，N+M=L 。  

然後證明的是， 對於有N個負數的數組， 這N個負數可以任意選擇。  這樣題目就變成了求解這個數組進行無限次操作，所能得到最小的負數的個數.

然後bfs 直接上。

那個性質很好證明。 不說了

 

C. Yaroslav and Sequencetime limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Yaroslav has an array, consisting of (2·n - 1) integers. In a single operation Yaroslav can change the sign of exactly n elements in the array. In other words, in one operation Yaroslav can select exactly n array elements, and multiply each of them by -1.

Yaroslav is now wondering: what maximum sum of array elements can be obtained if it is allowed to perform any number of described operations?

Help Yaroslav.

Input

The first line contains an integer n (2 ≤ n ≤ 100). The second line contains (2·n - 1) integers — the array elements. The array elements do not exceed 1000 in their absolute value.

Output

In a single line print the answer to the problem — the maximum sum that Yaroslav can get.

Sample test(s)input

2
50 50 50

output

150

input

2
-1 -100 -1

output

100

Note

In the first sample you do not need to change anything. The sum of elements equals 150.

In the second sample you need to change the sign of the first two elements. Then we get the sum of the elements equal to 100.

 

#include <stdio.h>#include <algorithm>#include <string.h>#include <iostream>using namespace std;int que[10010];int g[220];int mark[220];int main(){    int n;    scanf("%d",&n);    int cnt=0;    for(int i=0;i<2*n-1;i++)    {        scanf("%d",&g[i]);        if(g[i]<0) cnt++;    }    int qf=1,qd=0;    que[0]=cnt;    mark[cnt]=1;    while(qf>qd)    {        int cur=que[qd++];        for(int i=0;i<=cur;i++)        {            int tmp=n-i;            if(i<=n && tmp<=2*n-1-cur)            {                int nwnode=cur-i+tmp;                if(mark[nwnode]==0)                {                    mark[nwnode]=1;                    que[qf++]=nwnode;                }            }        }    }    int tmp;    for(int i=0;i<=200;i++)        if(mark[i]==1)         {            tmp=i;            break;        }    int ans=0;    for(int i=0;i<2*n-1;i++)    {        if(g[i]<0)        {            g[i]*=-1;        }        ans += g[i];    }    sort(g,g+2*n-1);    int sum=0;    for(int i=0;i<tmp;i++)        sum+=g[i];    printf("%d",ans-2*sum);    return 0;}