2018SDIBT_国庆个人第二场

Posted 2021-01-10 一只小毛球

C.1038C Gambling

C. Gambling

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Two players A and B have a list of $\mathrm{nn integers\; each.\; They\; both\; want\; to\; maximize\; the\; subtraction\; between\; their\; score\; and\; their\; opponent‘s\; score.}$

In one turn, a player can either add to his score any element from his list (assuming his list is not empty), the element is removed from the list afterward. Or remove an element from his opponent‘s list (assuming his opponent‘s list is not empty).

Note, that in case there are equal elements in the list only one of them will be affected in the operations above. For example, if there are elements $\{1,2,2,3\}\{1,2,2,3\} in\; a\; list\; and\; you\; decided\; to\; choose 22 for\; the\; next\; turn,\; only\; a\; single\; instance\; of 22 will\; be\; deleted\; (and\; added\; to\; the\; score,\; if\; necessary).$

The player A starts the game and the game stops when both lists are empty. Find the difference between A‘s score and B‘s score at the end of the game, if both of the players are playing optimally.

Optimal play between two players means that both players choose the best possible strategy to achieve the best possible outcome for themselves. In this problem, it means that each player, each time makes a move, which maximizes the final difference between his score and his opponent‘s score, knowing that the opponent is doing the same.

Input

The first line of input contains an integer $\mathrm{nn (1\le n\le 1000001\le n\le 100000) —\; the\; sizes\; of\; the\; list.}$

The second line contains $\mathrm{nn integers aiai (1\le ai\le 1061\le ai\le 106),\; describing\; the\; list\; of\; the\; player\; A,\; who\; starts\; the\; game.}$

The third line contains $\mathrm{nn integers bibi (1\le bi\le 1061\le bi\le 106),\; describing\; the\; list\; of\; the\; player\; B.}$

Output

Output the difference between A‘s score and B‘s score ($\mathrm{A-BA-B)\; if\; both\; of\; them\; are\; playing\; optimally.}$

Examples

input

Copy

2
1 4
5 1

output

Copy

0

input

Copy

3
100 100 100
100 100 100

output

Copy

0

input

Copy

2
2 1
5 6

output

Copy

-3

Note

In the first example, the game could have gone as follows:

• A removes $\mathrm{55 from\; B‘s\; list.}$
• B removes $\mathrm{44 from\; A‘s\; list.}$
• A takes his $11.$
• B takes his $11.$

Hence, A‘s score is $\mathrm{11,\; B‘s\; score\; is 11 and\; difference\; is 00.}$

There is also another optimal way of playing:

• A removes $\mathrm{55 from\; B‘s\; list.}$
• B removes $\mathrm{44 from\; A‘s\; list.}$
• A removes $\mathrm{11 from\; B‘s\; list.}$
• B removes $\mathrm{11 from\; A‘s\; list.}$

The difference in the scores is still $00.$

In the second example, irrespective of the moves the players make, they will end up with the same number of numbers added to their score, so the difference will be $00.$

$\mathrm{大致题意：有两个人a,b,有两个数组，每个人可以从自己的数组选一个数当作自己的积分，这个数被除去，或者从别人的数组里除去一个数，双方都会进行这个操作，当双方数组里的没有数字的时候，游戏结束，尽可能使得自己的积分最多，最后输出两个人最终积分的差}$

$\mathrm{分析：先给两个数组从小到大排个序，使用双指针，另i=n,j=n;}$

$\mathrm{如果a[i]>b[j]时，}$

$\mathrm{如果轮到a来选，那么a的积分ans1+=a[i],i--;}$

$\mathrm{如果轮到b来选，那么b将a[i]从数组a里除去，i--;}$

$\mathrm{如果a[i]<b[j]时,}$

$\mathrm{如果轮到a来选，那么a将b[j]从数组b里除去，j--;}$

$\mathrm{如果轮到b来选，那么b的积分ans2+=b[j]，j--;}$

$\mathrm{如果a[i]=b[j],}$

$\mathrm{跳过i,j，i--，j--}$

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,a[1000005],b[1000005];
 8     while(~scanf("%d",&n))
 9     {
10         for(int i=1;i<=n;i++)
11             scanf("%d",&a[i]);
12         for(int i=1;i<=n;i++)
13             scanf("%d",&b[i]);
14         sort(a+1,a+1+n);
15         sort(b+1,b+1+n);
16         a[0]=0;
17         b[0]=0;
18         long long k=0,ans1=0,ans2=0;
19         for(int i=n,j=n;i>=0&&j>=0;)
20         {
21             if(a[i]>b[j])
22             {
23                 if(k==0)//使用k来标记这轮轮到谁，0是a,1是b
24                 {
25                     ans1+=a[i];
26                     k=1;
27                     i--;
28                 }
29                 else
30                 {
31                     i--;
32                     k=0;
33                 }
34             }
35             else if(a[i]<b[j])
36             {
37                 if(k==1)
38                 {
39                     ans2+=b[j];
40                     k=0;
41                     j--;
42                 }
43                 else
44                 {
45                      k=1;
46                      j--;
47                 }
48             }
49             else if(a[i]==b[j])
50             {
51                 i--;
52                 j--;
53             }
54         }
55         printf("%lld\n",ans1-ans2);
56     }
57     return 0;
58 }