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  • hdu 1452 Happy 2004 (快速幂+取模乘法逆元)

    Problem Description
    Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).
    Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.
     
    Input
    The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000). 
    A test case of X = 0 indicates the end of input, and should not be processed.
     
    Output
    For each test case, in a separate line, please output the result of S modulo 29.
     
    Sample Input
    1 10000 0
     
    Sample Output
    6 10

       题意:找2014^x%29的值。

       快速幂和取模乘法逆元。乘法逆元:x=(1/b)%m. 求x的值。x*b=k*m+1,即k去最小正整数时,x也为整数。即b能被k*m+1整除。

      

     1 #include<cstdio>
     2 #include<cstring>
     3 using namespace std;
     4 int f1(int x,int y)
     5 {
     6     int ans=1;
     7     while (y)
     8     {
     9         if (y&1) ans=ans*x%29;
    10         x=x*x%29;
    11         y>>=1;
    12     }
    13     return ans;
    14 }
    15 int f2(int x)
    16 {
    17     int i=1;
    18     while (i)
    19     {
    20         if ((29*i+1)%x==0) break;
    21         i++;
    22     }
    23     return (29*i+1)/x;
    24 }
    25 int main()
    26 {
    27     int x,a,b,c;
    28     while (~scanf("%d",&x))
    29     {
    30         if (!x) break;
    31         a=(f1(2,2*x+1)-1)%29;
    32         c=(((f1(3,x+1)-1)%29)*f2(2))%29;
    33         b=(((f1(22,x+1)-1)%29)*f2(21))%29;
    34         printf("%d
    ",(a*b*c)%29);
    35     }
    36     return 0;
    37 }
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  • 原文地址:https://www.cnblogs.com/pblr/p/4746578.html
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