[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 v₁ and v₂ 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：

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

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
",casi,sum);
71 }
72     return 0;
73 }
74

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

Problem A. Minimum Scalar Product

排序不等式的证明

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