hdu 5933 ArcSoft's Office Rearrangement

Problem Description

ArcSoft, Inc. is a leading global professional computer photography and computer vision technology company.

There are N working blocks in ArcSoft company, which form a straight line. The CEO of ArcSoft thinks that every block should have equal number of employees, so he wants to re-arrange the current blocks into K new blocks by the following two operations:

- merge two neighbor blocks into a new block, and the new block's size is the sum of two old blocks'.
- split one block into two new blocks, and you can assign the size of each block, but the sum should be equal to the old block.

Now the CEO wants to know the minimum operations to re-arrange current blocks into K block with equal size, please help him.

Input

First line contains an integer T, which indicates the number of test cases.

Every test case begins with one line which two integers N and K, which is the number of old blocks and new blocks.

The second line contains N numbers a1, a2, ⋯, aN, indicating the size of current blocks.

Limits
1≤T≤100
1≤N≤105
1≤K≤105
1≤ai≤105

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum operations.

If the CEO can't re-arrange K new blocks with equal size, y equals -1.

Sample Input
3
1 3
14
3 1
2 3 4
3 6
1 2 3

Sample Output

Case #1: -1
Case #2: 2

Case #3: 3

思路：求新格子的大小new，然后从头到尾扫描一遍，大于new的就拆，小于的就和，就行了。有点小细节就是一定要用longlong当10^5个10^5大小的格子合成一个格子就会超int了。

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
long long int a[100005],sum,v,tep;
int main()
{
    int n,T,m;
    cin>>T;
    for(int t=1; t<=T; t++)
    {
        scanf("%d %d",&n,&m);
        tep=sum=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        v=sum/m;
        for(int i=1;i<=n;)
        {
            if(a[i]<v){
                a[i+1]+=a[i];
                i++;
                tep++;
            }else{
                if(a[i]>v){
                    a[i]=a[i]-v;
                    tep++;
                }else
                i++;
            }
        }
        if(sum%m)
        cout<<"Case #"<<t<<": "<<"-1"<<endl;
        else
        cout<<"Case #"<<t<<": "<<tep<<endl;
    }
    return 0;
}