「UVA12004」 Bubble Sort 解题报告
UVA12004 Bubble Sort
Check the following code which counts the number of swaps of bubble sort.
int findSwaps( int n, int a[] )
{
int count = 0, i, j, temp, b[100000];
for( i = 0; i < n; i++ ) {
b[i] = a[i];
}
for( i = 0; i < n; i++ ) {
for( j = 0; j < n - 1; j++ ) {
if( b[j] > b[j+1] ) {
temp = b[j];
b[j] = b[j+1];
b[j+1] = temp;
count++;
}
}
}
return count;
}
You have to find the average value of ’count’ in the given code if we run findSwaps()
infinitely many times using constant ’n’ and each time some random integers (from 1 to n) are given in array a[]
. You can assume that the input integers in array a[]
are distinct.
Input
Input starts with an integer T (≤ 1000), denoting the number of test cases. Each test case contains an integer n (1 ≤ n ≤ 105) in a single line.
Output
For each case, print the case number and the desired result. If the result is an integer, print it. Otherwise print it in ‘p/q’ form, where p and q are relative prime.
Sample Input
2
1
2
Sample Output
Case 1: 0
Case 2: 1/2
思路
一句话题意:求长度为n的排列的期望逆序对数。
很简单,\(f(n)=f(n-1)+\frac{n-1}2=\frac{n\times(n-1)}4,f(1)=0\)。
为什么呢?假设把\(n\)插入长度\((n-1)\)的排列,有\(n\)种方法。期望增加的逆序对数就是\(\frac{1+2+...n-1}n=\frac{n\times (n-1)}{2n}=\frac{n-1}2\)
所以\(f(n)=f(n-1)+\frac{n-1}2\)
很简单吧?别忘了开long long
代码
#include<bits/stdc++.h>
using namespace std;
#define LL long long
int T, i;
LL n;
int main(){
scanf( "%d", &T );
for ( int i = 1; i <= T; ++i ){
scanf( "%lld", &n );
n = n * ( n - 1 ) / 2;
if ( n & 1 ) printf( "Case %d: %lld/2\n", i, n );
else printf( "Case %d: %lld\n", i, n / 2 );
}
return 0;
}