题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=5933
题目大意:有n个块,每个块大小为ai,现在要变成k个块,每个块大小一样,现在有两个操作:把一个块分割成任意两个大小的块,或者将相邻的两个块合并,问最小操作数。
解题思路:遍历一遍每个块,然后分类讨论就好了……水题,注意long long 的坑。
/* ***********************************************
┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃ ┃ ┆
┆┃ ┻ ┃ ┆
┆┗━┓ 马 ┏━┛ ┆
┆ ┃ 勒 ┃ ┆
┆ ┃ 戈 ┗━━━┓ ┆
┆ ┃ 壁 ┣┓┆
┆ ┃ 的草泥马 ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;
#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q)
{
return q==0?p:Gcd(q,p%q);
}
SelfType Pow(SelfType p,SelfType q)
{
SelfType ans=1;
while(q)
{
if(q&1)ans=ans*p;
p=p*p;
q>>=1;
}
return ans;
}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
#define mem(x,v) memset(x,v,sizeof(x))
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read()
{
int ra,fh;
char rx;
rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();
if(rx=='-')fh=-1,rx=getchar();
while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();
return ra*fh;
}
LL a[100005];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
//ios::sync_with_stdio(0);
//cin.tie(0);
int t;
t = read();
int cas = 1;
while(t--)
{
LL n,k;
scanf("%I64d%I64d",&n,&k);
LL sum = 0;
for(int i=1; i<=n; i++) scanf("%I64d",&a[i]), sum += a[i];
if(sum%k!=0)
{
printf("Case #%d: -1\n",cas++);
continue;
}
LL ans=0;
LL avg = sum / k;
for(int i=1; i<=n; i++)
{
if(a[i]<avg)
{
if(a[i]+a[i+1]<=avg)
{
ans++;
a[i+1]=a[i]+a[i+1];
}
else
{
a[i+1]=a[i+1]-avg+a[i];
ans+=2;
}
}
else if(a[i]>avg)
{
int l=a[i]/avg;
int m=a[i]%avg;
if(m!=0)
{
a[i+1]+=m;
ans+=2;
}
ans+=l-1;
}
}
printf("Case #%d: %I64d\n",cas++,ans);
}
return 0;
}