HDU 5933 ArcSoft's Office Rearrangement (模拟)

ArcSoft’s Office Rearrangement

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 457 Accepted Submission(s): 213

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

Source
2016年中国大学生程序设计竞赛（杭州）

简单的模拟题，但是细节要处理好。因为是按顺序合并或者切开，将和看做一个轴的话，终态需要切的位置是固定的，初始状态也有一定的已经切好的位置，如果两个状态的某些切好的位置是一致的那么就不需要动他了，其他位置直接进行合并重新切即可。

#include "cstring"#include "cstdio"#include "string.h"#include "iostream"using namespace std;#define MAX 100005long long a[MAX];int main(){    int t;    scanf("%d",&t);    int cas=1;    while(t--)    {        long long n,k;        scanf("%lld%lld",&n,&k);        long long sum=0;        for(int i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            sum+=a[i];        }        if(sum%k!=0)        {            printf("Case #%d: -1\n",cas++);            continue;        }        long long cnt=0;        long long interval=sum/k;        long long temp=0;        for(int i=1;i<=n;i++)        {            long long old=temp;            temp+=a[i];            if(old!=0)                cnt++;            if(temp<interval)            {                continue;            }            if(temp%interval==0)            {                long long now=temp/interval;                cnt+=now-1;                temp%=interval;                continue;            }            if(temp>interval)            {                long long now=temp/interval;                cnt+=now;                temp%=interval;                continue;            }        }        printf("Case #%d: %lld\n",cas++,cnt);    }}