【2015-2016 ACM-ICPC, NEERC, Southern Subregional Contest I】【水题】Lottery 均分气球最小修改数
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#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=105,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int n,k,x;int a[N];int main(){while(~scanf("%d%d",&n,&k)){MS(a,0);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;}/*【题意】有k个人,n个球。(1<=k<=n<=100,且n是k的倍数)每个球初始都有一个所属者,最后我们想使得每个人都拥有相同数量的球。问你最少的对求所属者的修改量是多少。【类型】水题【分析】直接求(∑abs(每个人拥有球量-ave))/2即可*/
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