码迷,mamicode.com
首页 > 其他好文 > 详细

“玲珑杯”ACM比赛 Round #1 A -- Absolute Defeat

时间:2016-08-20 17:35:14      阅读:122      评论:0      收藏:0      [点我收藏+]

标签:

DESCRIPTION
Eric has an array of integers 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 mm. Help him find the minimum number of operations needed.
INPUT
There are multiple test cases. The first line of input contains an integer TT, indicating the number of test cases. For each test case: The first line contains three integers nn, mm and kk (1n105,1kn,1m104)(1≤n≤105,1≤k≤n,1≤m≤104). The second line contains nn integers a1,a2,...,ana1,a2,...,an (1ai104)(1≤ai≤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

题目链接:http://ifrog.cc/acm/problem/1014?contest=1001&no=0

*************************************

错误代码:

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<time.h>
 7 #include<stack>
 8 using namespace std;
 9 #define N 120000
10 #define INF 0x3f3f3f3f
11 
12 int a[N];
13 
14 int main()
15 {
16     int T,n,m,k,i;
17 
18     scanf("%d", &T);
19 
20     while(T--)
21     {
22         scanf("%d %d %d", &n,&m,&k);
23 
24         for(i=0; i<n; i++)
25             scanf("%d", &a[i]);
26 
27         sort(a,a+n);
28 
29         int i=0,sum=0;
30         while(i<n)
31         {
32             if(m-a[i]>0)
33                 sum+=m-a[i];
34             i+=k;
35         }
36         printf("%d\n",sum);
37     }
38 
39     return 0;
40 }

 

 

 

“玲珑杯”ACM比赛 Round #1 A -- Absolute Defeat

标签:

原文地址:http://www.cnblogs.com/weiyuan/p/5790673.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!