「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

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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;
}