Problem DescriptionThe Inversion number of a given number sequence A1, A2, ..., the number of pairs (AI, aj) that SA Tisfy i < J and Ai > aj.
For a given sequence of numbers a1, A2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we'll Obtain another sequence. There is totally n such sequences as the following:
A1, A2, ..., An-1, an (where m = 0-the initial seqence)
A2, A3, ..., an, A1 (where m = 1)
A3, A4, ..., an, A1, A2 (where m = 2)
...
An, A1, A2, ..., an-1 (where m = n-1)
You is asked to write a program to find the minimum inversion number out of the above sequences. Inputthe input consists of a number of test cases. Each case consists of the lines:the first line contains a positive integer n (n <= 5000); The next line contains a permutation of the n integers from 0 to n-1. Outputfor each case, output the minimum inversion number to a single line. Sample Input101 3 6 9 0 8 5 7 4 2 sample Output16 roughly test instructions: one by 0. N-1 sequence, each time you can move the first element of the team to the end of the team;
The number of the smallest inverse pairs in the n sequence formed; The problem-solving idea: Violence law: When the a[0] moves to the end of the team, it will reduce the a[0] reverse order will increase (n-1)-a[0] in reverse order; then only the initial reverse logarithm is required, the rest can be deduced;
1#include <cstdio>2 using namespacestd;3 intMain () {4 intn,a[ the+5],t;5 while(~SCANF ("%d",&N)) {6 for(intI=0; i<n;i++) scanf ("%d",&a[i]);7t=0;8 for(intI=0; i<n;i++)9 for(intj=i+1; j<n;j++)Ten if(A[i]>a[j]) t++; One intmin=T; A for(intI=0; i<n;i++){ -t=t-a[i]+ (n1)-A[i]; - if(t<min) min=T; the } -printf"%d\n", min); - } -}
HDU 1394-minimum Inversion number
