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  • Gold Coins(2000)

    Gold Coins

    Description

    The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the next three days (the fourth, fifth, and sixth days of service), the knight receives three gold coins. On each of the next four days (the seventh, eighth, ninth, and tenth days of service), the knight receives four gold coins. This pattern of payments will continue indefinitely: after receiving N gold coins on each of N consecutive days, the knight will receive N+1 gold coins on each of the next N+1 consecutive days, where N is any positive integer. 

    Your program will determine the total number of gold coins paid to the knight in any given number of days (starting from Day 1). 

    Input

    The input contains at least one, but no more than 21 lines. Each line of the input file (except the last one) contains data for one test case of the problem, consisting of exactly one integer (in the range 1..10000), representing the number of days. The end of the input is signaled by a line containing the number 0.

    Output

    There is exactly one line of output for each test case. This line contains the number of days from the corresponding line of input, followed by one blank space and the total number of gold coins paid to the knight in the given number of days, starting with Day 1.

    Sample Input

    10 
    11 
    15 
    16 
    100 
    10000 
    1000 
    21 
    22 
    0 

    Sample Output

    10 30 
    6 14 
    7 18 
    11 35 
    15 55 
    16 61 
    100 945 
    10000 942820 
    1000 29820 
    21 91 
    22 98 

    Source

    AC code:

    #include<stdio.h>
    #include
    <math.h>
    #include
    <string.h>

    int main()
    {
         
    int n;
         
    while((scanf("%d",&n)!=EOF)&&n)
         {
              
    int i,j,t;
              
    long int sum=0;
              
    for(i=1; ; i++)
              {
                   
    if(i*(i+1)/2>=n)
                   {
                        
    break;
                      }
                 }
                 
    for(j=1; j<i; j++)
                 {
                      sum
    +=j*j;
                    }
                 t
    =i*(i+1)/2-i+1;
                 t
    =n-t+1;
                 sum
    +=t*i;
                    printf(
    "%d %ld\n",n,sum);
            }
         
         
    return 0;

    } 

    作者:cpoint
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
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  • 原文地址:https://www.cnblogs.com/cpoint/p/2017592.html
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