LotteryCrawling in process... Crawling failed Time Limit:2000MS Memory Limit:524288KB 64bit IO Format:%I64d & %I64u
Description
Input
Output
Sample Input
Sample Output
Hint
Description
Today Berland holds a lottery with a prize — a huge sum of money! There are k persons, who attend the lottery. Each of them will receive a unique integer from 1 to k.
The organizers bought n balls to organize the lottery, each of them is painted some color, the colors are numbered from 1 to k. A ball of color c corresponds to the participant with the same number. The organizers will randomly choose one ball — and the winner will be the person whose color will be chosen!
Five hours before the start of the lottery the organizers realized that for the lottery to be fair there must be an equal number of balls of each of k colors. This will ensure that the chances of winning are equal for all the participants.
You have to find the minimum number of balls that you need to repaint to make the lottery fair. A ball can be repainted to any of the k colors.
Input
The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 100) — the number of balls and the number of participants. It is guaranteed that n is evenly divisible by k.
The second line of the input contains space-separated sequence of n positive integers ci(1 ≤ ci ≤ k), where ci means the original color of the i-th ball.
Output
In the single line of the output print a single integer — the minimum number of balls to repaint to make number of balls of each color equal.
Sample Input
4 2
2 1 2 2
1
8 4
1 2 1 1 1 4 1 4
3
Sample Output
Hint
In the first example the organizers need to repaint any ball of color 2 to the color 1.
In the second example the organizers need to repaint one ball of color 1 to the color 2 and two balls of the color 1 to the color 3.
/* 大水题，最后才看到真是可惜啊 k个人n个球，n是k的倍数 每个人都有初始的球数，让你球使每个人的球数都相等的，最少要改变多少球数 */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #define N 105 using namespace std; int n,k,x; int a[N]; int main() { while(~scanf("%d%d",&n,&k)) { memset(a,0,sizeof a); int ave=n/k; for(int i=1;i<=n;i++) { scanf("%d",&x); ++a[x]; } int ans=0; for(int i=1;i<=k;i++)ans+=abs(a[i]-ave); printf("%d\n",ans>>1); } return 0; }
