[Google Code Jam (Round 1A 2008) ] A. Minimum Scalar Product

Problem A. Minimum Scalar Product

 

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Problem

You are given two vectors v₁=(x₁,x₂,...,x_n) and v₂=(y₁,y₂,...,y_n). The scalar product of these vectors is a single number, calculated as x₁y₁+x₂y₂+...+x_ny_n.

Suppose you are allowed to permute the coordinates of each vector as you wish. Choose two permutations such that the scalar product of your two new vectors is the smallest possible, and output that minimum scalar product.

Input

The first line of the input file contains integer number T - the number of test cases. For each test case, the first line contains integer number n. The next two lines contain n integers each, giving the coordinates of v ₁ and v ₂ respectively.

 

Output

For each test case, output a line

Case #X: Y

where X is the test case number, starting from 1, and Y is the minimum scalar product of all permutations of the two given vectors.

 

Limits

 

Small dataset

T = 1000
1 ≤ n ≤ 8
-1000 ≤ x_i, y_i ≤ 1000

Large dataset

T = 10
100 ≤ n ≤ 800
-100000 ≤ x_i, y_i ≤ 100000

Sample

Input	Output
2 3 1 3 -5 -2 4 1 5 1 2 3 4 5 1 0 1 0 1	Case #1: -25 Case #2: 6

题解：排序不等式的应用。

 

排序不等式公式

0<a1<a2<a3......<an

0<b1<b2<b3......<bn

an×bn+an-1×bn-1+......+a1×b1>=乱序和>=a1×bn+a2×bn-1+......+an×b1

(注：n，n-1，n-2等，均为角标）

 

顺序不等式基本形式：

排序不等式的证明

分析法

要证

只需证

根据基本不等式

只需证

∴原结论正确

 

代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 int i,j,n,m;

 5 

 6 long long    a[1000],b[1000];

 7 

 8 long long sum;

 9 

10 int 

11 pre()

12 {

13     memset(a,0,sizeof(a));

14     memset(b,0,sizeof(b));

15     sum=0;

16     return 0;

17 }

18 

19 int 

20 init()

21 {

22     scanf("%d",&n);

23     for(i=1;i<=n;i++)

24     scanf("%lld",&a[i]);

25     for(i=1;i<=n;i++)

26     scanf("%lld",&b[i]);

27     

28     return 0;

29 }

30 

31 void

32 qsort(long long a[],int head,int tail)

33 {

34     int i,j;

35     long long x;

36     i=head;j=tail;

37     x=a[head];

38     

39     while(i<j)

40     {

41         while((i<j)&(a[j]>=x)) j--;

42         a[i]=a[j];

43         while((i<j)&(a[i]<=x)) i++;

44         a[j]=a[i];

45     }

46     a[i]=x;

47     

48     if(head<(i-1)) qsort(a,head,i-1);

49     if((i+1)<tail) qsort(a,i+1,tail);

50 }

51 

52 

53 int 

54 main()

55 {

56     int casi,cas;

57     freopen("1.in","r",stdin);

58     freopen("1.out","w",stdout);

59     scanf("%d",&cas);

60 for(casi=1;casi<=cas;casi++)

61 {

62     pre();

63     init();

64     qsort(a,1,n);

65     qsort(b,1,n);

66     

67     for(i=1;i<=n;i++)

68     sum+=a[i]*b[n-i+1];

69     

70     printf("Case #%d: %lld\n",casi,sum);

71 }

72     return 0;

73 }

74