zoukankan      html  css  js  c++  java
  • CF Dima and To-do List

    B. Dima and To-do List
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.

    Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete k - 1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.

    Overall, Dima has n tasks to do, each task has a unique number from 1 to n. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.

    Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.

    Input

    The first line of the input contains two integers n, k (1 ≤ k ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103), where ai is the power Inna tells Dima off with if she is present in the room while he is doing the i-th task.

    It is guaranteed that n is divisible by k.

    Output

    In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.

    Sample test(s)
    input
    6 2
    3 2 1 6 5 4
    output
    1
    input
    10 5
    1 3 5 7 9 9 4 1 8 5
    output
    3
    Note

    Explanation of the first example.

    If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as k = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.

    Explanation of the second example.

    In the second example k = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.

     1 #include<stdio.h>
     2 #include<stdlib.h>
     3 #include<string.h>
     4 
     5 int a[100003];
     6 void solve(int n,int m)
     7 {
     8     int i,j,k,Min=100000110,wz=1;
     9     for(i=1;i<=m;i++)
    10     {
    11         for(j=i,k=0;j<=n;j=j+m)
    12         k=k+a[j];
    13         if(Min>k)
    14         {
    15             Min=k;
    16             wz=i;
    17         }
    18     }
    19     printf("%d
    ",wz);
    20 }
    21 int main()
    22 {
    23     int n,m,i;
    24     while(scanf("%d%d",&n,&m)>0)
    25     {
    26         for(i=1;i<=n;i++)
    27             scanf("%d",&a[i]);
    28         if(m==1)
    29         {
    30             printf("1
    ");
    31             continue;
    32         }
    33         solve(n,m);
    34     }
    35     return 0;
    36 }
  • 相关阅读:
    postgresql批量插入copy_from()的使用
    Fabric SSH链接时关于找不到主机的问题
    Python多线程获取返回值
    网页正文提取,降噪的实现(readability/Document)
    HTML标签参考手册
    javascript获取当前日期和时间
    2017易观OLAP算法大赛
    Apache DolphinScheduler 诞生记
    Apache DolphinScheduler(海豚调度)
    开源分布式工作流任务调度系统Easy Scheduler Release 1.0.2发布
  • 原文地址:https://www.cnblogs.com/tom987690183/p/3442409.html
Copyright © 2011-2022 走看看