UESTC 1817 Complete Building the Houses

Complete Building the Houses

Time Limit: 2000MS

Memory Limit: 65535KB

64bit IO Format: %lld & %llu

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Description

Bear has a large, empty ground for him to build a home. He decides to build a row of houses, one after another, say n in total.

The houses are designed with different height. Bear has m workers in total, and the workers must work side by side. So at a time bear can choose some continuous houses, no more than m, and add their heights by one, this takes one day to finish.

Given the designed height for each house, what is the minimum number of days after which all the houses’ heights are no less than the original design?

Input

The first line of input contains a number T, indicating the number of test cases. (T<=50)

For each case, the first line contains two integers n and m: the number of houses and the number of workers. The next line comes with n non-negative numbers, they are the heights of the houses from left to right. (1<=n, m<=100,000, each number will be less than 1,000,000,000)

Output

For each case, output “Case #i: “ first. (i is the number of the test case, from 1 to T). Then output the days when bear’s home can be built.

Sample Input

2
3 3
1 2 3
3 3
3 2 1

Sample Output

Case #1: 3
Case #2: 3

Source

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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int sum[110000];
long long int ans=0,d;

int main()
{
    int T,m,n,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        ans=0;  int t;
        scanf("%d%d",&n,&m);
        memset(sum,0,sizeof(sum));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&t);
            if(i-1>=0) sum+=sum[i-1];
            if(t-sum>0)
            {
                d=t-sum;
                sum+=d;
                ans+=d;
                if(i+m<n)
                {
                    sum[i+m]-=d;
                }
            }
        }
        printf("Case #%d: %lld\n",cas++,ans);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )