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  • hdu 1012:u Calculate e(数学题,水题)

    u Calculate e

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 28686    Accepted Submission(s): 12762


    Problem Description
    A simple mathematical formula for e is



    where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
     
    Output
    Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
     
    Sample Output
    n e
    - -----------
    0 1
    1 2
    2 2.5
    3 2.666666667
    4 2.708333333
     
    Source
     
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      水题
      切一道水题,放松下心情。
      这道题没有输入,只有输出。
      前3组数据由于输出格式不统一,直接输出即可。后面的数可用迭代思路求得,不用从头重新计算了。
      代码:
     1 #include <stdio.h>
     2 int main()
     3 {
     4     int i;
     5     printf("n e
    ");    
     6     printf("- -----------
    ");
     7     printf("0 1
    ");    //没有统一格式,提前输出。
     8     printf("1 2
    ");
     9     printf("2 2.5
    ");
    10     double ans = 1;
    11     double t = 1;
    12     for(i=1;i<=9;i++){
    13         t = 1.0/i*t;    //迭代计算
    14         ans += t;
    15         if(i<3) continue;
    16         printf("%d %.9lf
    ",i,ans);
    17     }
    18     return 0;
    19 }

    Freecode : www.cnblogs.com/yym2013

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  • 原文地址:https://www.cnblogs.com/yym2013/p/3695088.html
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