HDU5933ArcSoft's Office Rearrangement（模拟）

最新推荐文章于 2019-09-22 10:40:26 发布

吃货妍

最新推荐文章于 2019-09-22 10:40:26 发布

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解决一个关于办公室重组的问题，目标是将N个工作区块重新分配为K个大小相同的新区块，通过合并或拆分区块来实现。本文提供了一个算法解决方案，并附带源代码。

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ArcSoft’s Office Rearrangement

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 422 Accepted Submission(s): 199

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
题意：把一个长度为n的数组，分成k个相等区间，（可以合并相邻区间或者将一个区间拆开）输出最少操作次数，不能输出-1。
题解:模拟题,注意数据范围。
代码:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+10;
int T,n,k;
ll a[N];
int solo(int n,int k)
{
    ll ans=0;
    for(int j=1;j<=n;j++)
    {
        scanf("%lld",&a[j]);
        ans+=a[j];
    }
    if(ans%k!=0)
        return -1;
        ll tmp=ans/k;
        int res=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]<tmp)
        {
            if(a[i]+a[i+1]<=tmp)
            {
                res++;
                a[i+1]+=a[i];
            }
            else
            {
                a[i+1]=a[i]+a[i+1]-tmp;
                res+=2;
            }
        }
        else
        {
            ll q=a[i]/tmp;
            ll p=a[i]%tmp;
            if(p!=0)
            {
                a[i+1]+=p;
                res+=2;
            }
            res+=q-1;
        }
    }
    return res;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        printf("Case #%d: ",i);
        scanf("%d%d",&n,&k);
        printf("%d\n",solo(n,k));
    }
    return 0;
}