“玲珑杯”ACM比赛 Round #1

Start Time：2016-08-20 13:00:00 End Time：2016-08-20 18:00:00 Refresh Time：2017-11-12 19:51:52 Public

A -- Absolute Defeat

Time Limit：2s Memory Limit：64MByte

Submissions：394Solved：119

DESCRIPTION

Eric has an array of integers ${\mathrm{a1,a2,...,ana1,a2,...,an.\; Every\; time,\; he\; can\; choose\; a\; contiguous\; subsequence\; of\; length\; kk\; and\; increase\; every\; integer\; in\; the\; contiguous\; subsequence\; by\; 11.}}^{}$

He wants the minimum value of the array is at least $\mathrm{mm.\; Help\; him\; find\; the\; minimum\; number\; of\; operations\; needed.}$

INPUT

There are multiple test cases. The first line of input contains an integer $\mathrm{TT,\; indicating\; the\; number\; of\; test\; cases.\; For\; each\; test\; case:\; The\; first\; line\; contains\; three\; integers\; nn,\; mm\; and\; kk\; (1\le n\le 105,1\le k\le n,1\le m\le 104)(1\le n\le 105,1\le k\le n,1\le m\le 104).\; The\; second\; line\; contains\; nn\; integers\; a1,a2,...,ana1,a2,...,an\; (1\le ai\le 104)(1\le ai\le 104).}$

 

OUTPUT

For each test case, output an integer denoting the minimum number of operations needed.

 

SAMPLE INPUT

3

2 2 2

1 1

5 1 4

1 2 3 4 5

4 10 3

1 2 3 4

SAMPLE OUTPUT

1

0

15

SOLUTION

“玲珑杯”ACM比赛 Round #1

 

【题意】：给你一个长度为n的序列，每次你只能选择连续的k个数字加1，问你要进行多少次操作才能使所有的a[i]都大于m。 

【分析】：直接枚举，看每个数是否>m,否的话给从它开始的k个数，每个数都加上m-a[i]。

【代码】：

#include <bits/stdc++.h>

using namespace std;
int a[1000000+10];
int main()
{
    int t,n,m,k;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>k;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        int ans=0;
        int tmp;
        for(int i=0;i<n;i++)
        {
            if(a[i]<m)
            {
                tmp=m-a[i];
                ans+=tmp;
                for(int j=i;j<i+k;j++) //连续长度k内元素都加上差值
                {
                    a[j]+=tmp;
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}