zoukankan      html  css  js  c++  java
  • 杭电_ACM_Number Sequence

    Problem Description
    A number sequence is defined as follows:

    f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

    Given A, B, and n, you are to calculate the value of f(n).
     
    Input
    The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
     
    Output
    For each test case, print the value of f(n) on a single line.
     
    Sample Input
    1 1 3
    1 2 10
    0 0 0
     
    Sample Output
    2
    5
    View Code
     1 #include <stdio.h>
     2 int main()
     3 {
     4     int n, A, B, a[100], T, i;
     5     while (scanf("%d %d %d", &A, &B, &n) != EOF)
     6     {
     7         if (n == 0 && A == 0 && B == 0)
     8             break;
     9         a[1] = 1;
    10         a[2] = 1;
    11         if (n == 1 || n == 2)
    12         {
    13             printf("%d\n", a[n]);
    14             continue;
    15         }
    16         if (A % 7 == 0 && B % 7 == 0)
    17         {
    18             printf("0\n");
    19             continue;
    20         }
    21         for (i = 3; i < 100; i++)
    22         {
    23             a[i] = (A * a[i - 1] + B * a[i - 2]) % 7;
    24             if (a[i] == 1 && a[i - 1] == 1)
    25             {
    26                 T = i - 2;
    27                 break;
    28             }
    29         }
    30         if (n % T == 0)
    31             n = T;
    32         else
    33             n %= T;
    34         printf("%d\n", a[n]);
    35     }
    36     return 0;
    37 }

    My process:

    firstly,

    View Code
     1 #include <stdio.h>
     2 int f(int A, int B, int n)
     3 {
     4     if (n == 1)
     5         return 1;
     6     else if (n == 2)
     7         return 1;
     8     else 
     9         return (A * f(A, B, n - 1) + B * f(A, B, n - 2)) % 7;
    10 }
    11 int main()
    12 {
    13     int n, A, B, result;
    14     while (scanf("%d %d %d", &A, &B, &n) != EOF)
    15     {
    16         A %= 7;
    17         B %= 7;
    18         if (n == 0 && A == 0 && B == 0)
    19             break;
    20         result = f(A, B, n);
    21         printf("%d\n", result);
    22     }
    23     return 0;
    24 }

    it's Runtime Error(STACK_OVERFLOW).

    secondly,

    View Code
     1 #include <stdio.h>
     2 int main()
     3 {
     4     int n, A, B, result, a, b;
     5     while (scanf("%d %d %d", &A, &B, &n) != EOF)
     6     {
     7         if (n == 0 && A == 0 && B == 0)
     8             break;
     9         a = 1;
    10         b = 1;
    11         A %= 7;
    12         B %= 7;
    13         if (n == 1 || n == 2)
    14         {
    15             printf("%d\n", a);
    16             continue;
    17         }
    18         for (int i = 3; i <= n; i++)
    19         {
    20             result = (A * b + B * a) % 7;
    21             a = b;
    22             b = result;
    23         }
    24         printf("%d\n", result);
    25     }
    26     return 0;
    27 }

    it's Time Limit Exceeded.

    thirdly,

    View Code
     1 #include <stdio.h>
     2 #define MAX 100000003
     3 int a[MAX];
     4 int main()
     5 {
     6     /**/
     7     int n, A, B, result;
     8     a[1] = 1;
     9     a[2] = 1;
    10     while (scanf("%d %d %d", &A, &B, &n) != EOF)
    11     {
    12         A %= 7;
    13         B %= 7;
    14         if (n == 0 && A == 0 && B == 0)
    15             break;
    16         if (n == 1 || n == 2)
    17         {
    18             printf("%d\n", a[n]);
    19             continue;
    20         }
    21         for (int i = 3; i <= n; i++)
    22         {
    23             a[i] = (A * a[i - 1] + B * a[i - 2]) % 7;
    24         }
    25         printf("%d\n", a[n]);
    26     }
    27     return 0;
    28 }

    it's Memory Limit Exceeded.

    ---------------------------

    Key points

    firstly, it's easy to get that it is circulatory. Because the result is zero to six, and if you want to get the continuing one. it will be emerged in 50 times, but if A, B are the multiple of 7, the resulte is zero when n >= 3.

    secondly, pay attention to details, if n is the mutiple of 7, you must set n to T.

    last but not least, for either question, you must have clear thought and pay attention to details.

     

     

  • 相关阅读:
    整合Django的信息显示框架messages framework
    django使用session来保存用户登录状态
    为窗体添加防机器人的验证机制
    模拟窗口类ModelForm的应用
    django基础窗口类的使用
    django实现利用mailgun进行收发邮件
    django中使用mysql数据库
    ubuntu安装dockers和images:dvwa
    JVM-对象
    JVM-运行时数据区
  • 原文地址:https://www.cnblogs.com/chuanlong/p/2747053.html
Copyright © 2011-2022 走看看