mod

问题 I need to perform a real mathematical modulo in C. It makes sense for me to allow negative numbers for the moduled argument, since my modular calculations can produce negative intermediate results, which must be put back into the least residue system. But it makes no sense to allow negative module, therefore i wrote unsigned int mod( int x, unsigned int m ) { int r = x % m; return r >= 0 ? r : r + m; } However calling such function with negative number and positive module printf("%u\n", mod(

问题 I need to perform a real mathematical modulo in C. It makes sense for me to allow negative numbers for the moduled argument, since my modular calculations can produce negative intermediate results, which must be put back into the least residue system. But it makes no sense to allow negative module, therefore i wrote unsigned int mod( int x, unsigned int m ) { int r = x % m; return r >= 0 ? r : r + m; } However calling such function with negative number and positive module printf("%u\n", mod(

问题 I write down a code which find out quotient after dividing two number but without using multiplication,division or mod operator. My code public int divide(int dividend, int divisor) { int diff=0,count=0; int fun_dividend=dividend; int fun_divisor=divisor; int abs_dividend=abs(dividend); int abs_divisor=abs(divisor); while(abs_dividend>=abs_divisor){ diff=abs_dividend-abs_divisor; abs_dividend=diff; count++; } if(fun_dividend<0 && fun_divisor<0){ return count; } else if(fun_divisor<0||fun

问题 I write down a code which find out quotient after dividing two number but without using multiplication,division or mod operator. My code public int divide(int dividend, int divisor) { int diff=0,count=0; int fun_dividend=dividend; int fun_divisor=divisor; int abs_dividend=abs(dividend); int abs_divisor=abs(divisor); while(abs_dividend>=abs_divisor){ diff=abs_dividend-abs_divisor; abs_dividend=diff; count++; } if(fun_dividend<0 && fun_divisor<0){ return count; } else if(fun_divisor<0||fun

问题 I write down a code which find out quotient after dividing two number but without using multiplication,division or mod operator. My code public int divide(int dividend, int divisor) { int diff=0,count=0; int fun_dividend=dividend; int fun_divisor=divisor; int abs_dividend=abs(dividend); int abs_divisor=abs(divisor); while(abs_dividend>=abs_divisor){ diff=abs_dividend-abs_divisor; abs_dividend=diff; count++; } if(fun_dividend<0 && fun_divisor<0){ return count; } else if(fun_divisor<0||fun

问题 I am trying to compute a^b^c mod p for some positive integers a,b,c,p. One possible (and obvious) way is to use fast modular exponentiation which will run in O(log(b^c))=clog(b) . While I don't mind the efficiency here, the obvious downfall of this method is that you need an explicit binary representation of b^c which in itself is already exponential. So the question for me is, if I can not represent b^c as a binary representation, is there a way I can compute a^b^c mod p from the binary

问题 I'm fairly new to angular2 and for the past few days I have been trying to create reusable form components using model driven forms So lets say we have a component componentA.component.ts @Component({ selector: 'common-a', template: ` <div [formGroup]="_metadataIdentifier"> <div class="form-group"> <label>Common A[1]</label> <div> <input type="text" formControlName="valueA1"> <small>Description 1</small> </div> <div class="form-group"> <label>Common A[2]</label> <div> <input type="text"

问题 Is there any algorithm to calculate (1^x + 2^x + 3^x + ... + n^x) mod 1000000007 ? Note: a^b is the b-th power of a. The constraints are 1 <= n <= 10^16, 1 <= x <= 1000 . So the value of N is very large. I can only solve for O(m log m) if m = 1000000007 . It is very slow because the time limit is 2 secs. Do you have any efficient algorithm? There was a comment that it might be duplicate of this question, but it is definitely different. 回答1: You can sum up the series 1**X + 2**X + ... + N**X

问题 I need a variable to point the array index, and like circle to return 0 when it arrive the end of array. I know that I can use if statement to judge, but I'm not sure it's will more quickly or not to use mod to achieve same function, can anyone give me some advise? int p=0; int arr[10]; void add_index(){ if(p==9) p=0; else p++; } Or int p=0; int arr[10]; void add_index(){ p=(p+1)%10; } 回答1: Once upon a time, definitely yes. These days, possibly no! I'll take the Intel Skylake as an example.

问题 I am working with mod_zeropush and i was working fine with 15.07 until i switched to the ejabberd version 16.03. I am using Ubuntu OS 14.04 Copied all beam files again to ebin directory it iis last error === 2016-05-12 15:40:08.706 [error] <0.499.0>@ejabberd_hooks:run1:332 {undef,[{xml,get_tag_attr_s,[<<"type">>,{xmlel,<<"message">>,[{<<"type">>, <<"chat">>},{<<"image">>,<<"http://example.com/web-services/user_images /personal_image/support_team_01.03.2016.png">>},{<<"id">>, <<