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**本原勾股数组(简写为PPT)是多个安慕希组(a,b,c),在那之中a,b,c没有公因数,且知足a^二

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端详直击:http://g.happypingpang.cn/topic_detail.php?liaoba=1959&topic=4b2ea190-1dd5-11b2-9bdc-7d103c4e7c69

  • b^贰 = c^2。举个例子下边是1项本原勾股数组:**

varhash = hex_md5("123dafd");

(3,4, 5),(5,12,13),(8,15,17),(7,24,25),(9,40,41),(11,60,61),(28,45,56),(33,56,65)。

alert(hash)

由这些短表轻便获得一些定论,举个例子,仿佛a与b奇偶性不相同且c总是奇数。

/*

注解如下:

* A JavaScript implementation of the RSA Data Security, Inc. MD5 Message

**若a与b都以偶数,则c也是偶数,意味着a,b,c有公因数二,所以安慕希组不是原本的,其次,若a,b都以奇数,那么c必然是偶数,那样假诺a=二x 一,b=二y 壹,c=贰z,代入a^2

* Digest Algorithm, as defined in RFC 1321.

  • b^贰 = c^贰,化简后得贰x^二 二x 二y^二 二y 一 = 贰z,明显不树立。所以,考虑到a,b的调换性,大家得出a是奇数,b是偶数,a,b,c未有公因数。**

* Version 2.1 Copyright (C) Paul Johnston 1999 - 2002.

紧接着观望如下,若是(a,b,c)是本原勾股数组,则可进展因数分解:

* Other contributors: Greg Holt, Andrew Kepert, Ydnar, Lostinet

a^2 = c^2 – b^2 = (c - b) * (c b)

* Distributed under the BSD License

上边为比方:

* See for more info.

3^2 = 5^2 - 4^2 = (5-4)(5 4) = 1 * 9,

*/

15^2 = 17^2 - 8^2 = (17-8)(17 8) = 9 * 25,

/*

21^2 = 29^2 - 20^2 = (29-20)(29 20) = 9 * 49。

* Configurable variables. You may need to tweak these to be compatible with

咱俩发掘仿佛c - b与c b本身总是平方数,怎么样注明呢?

* the server-side, but the defaults work in most cases.

由前边观察,c – b与c b仿佛未有公因数,注脚如下:

*/

**若是正整数d是c-b与c b的公因数,则d也整除(c

var hexcase = 0;  /* hex output format. 0 - lowercase; 1 - uppercase        */

  • b) ( c b) = 2c 和 (c b) - (c - b) = 二b,可是b和c未有公因数,所以d = 一或d = 2,可是d也整除(c - b)* (c b) = a^2,因为a是奇数,所以只好是d = 壹,所以c - b与c b未有公因数。**

var b64pad  = ""; /* base-64 pad character. "=" for strict RFC compliance  */

近日我们明白c-b与c b无公因数并且相乘为平方数,所以c-b与c b本身都以平方数,记:

var chrsz  = 8;  /* bits per input character. 8 - ASCII; 16 - Unicode      */

c b = s^二, c – b = t^2, 个中s> t≥壹是从未有过公因数的奇数。所以解为:

/*

b = (s^2 - t^2)/ 2,

* These are the functions you'll usually want to call

c = (s^2 t^2)/ 2,

* They take string arguments and return either hex or base-64 encoded strings

a = s * t.

*/

故此得出勾股数组定理:

function hex_md5(s){ return binl2hex(core_md5(str2binl(s), s.length * chrsz));}

各样本原勾股数组(a,b,c)(个中a为奇数,b为偶数)都可从如下公式得出:

金沙澳门官网,function b64_md5(s){ return binl2b64(core_md5(str2binl(s), s.length * chrsz));}

a = s * t,**b = (s^2 - t^2)/ 2,  c = (s^2 t^2)/ 2。**

function str_md5(s){ return binl2str(core_md5(str2binl(s), s.length * chrsz));}

中间a>t≥壹是随便未有公因数的奇数。

function hex_hmac_md5(key, data) { return binl2hex(core_hmac_md5(key, data)); }

例如取t=1,则安慕希组(s,  (s^二 - 一)/ 二, (s^贰 一)/ 二),它的b与c值仅相差一.

function b64_hmac_md5(key, data) { return binl2b64(core_hmac_md5(key, data)); }

 

function str_hmac_md5(key, data) { return binl2str(core_hmac_md5(key, data)); }

后天恰好做了1道有关的题:

/*

难题链接:codeforces 707C. Pythagorean Triples

* Perform a simple self-test to see if the VM is working

题意:给出二个数,构造勾股数。

*/

题解:

function md5_vm_test()

a≤2时无解;

{

a为奇数时,间接用地点取t = 一的公式得b = (a^贰 - 1)/ 二, c = (a^二 一)/ 2;

return hex_md5("abc") == "900150983cd24fb0d6963f7d28e17f72";

a为偶数时,当a = 肆 *d时,直接能够套勾股数(三,4,5),从而b= 三 *d, c = 5 * d,

}

                     对于其他的偶数a,能够a = a/ 2, 将a化为奇数,再一贯用地方公式就可以.

/*

 1 #include<cstdio>
 2 int main(){
 3     long long a;
 4     scanf("%I64d", &a);
 5     if(a == 1 || a == 2)
 6         printf("-1n");
 7     else if(a % 4 == 0){
 8         printf("%I64d %I64dn", a/4*3, a/4*5);
 9     }
10     else if(a % 2 == 0){
11         a /= 2;
12         printf("%I64d %I64dn", (a*a-1)/2*2, (a*a 1)/2*2);
13     }
14     else
15         printf("%I64d %I64dn", (a*a-1)/2, (a*a 1)/2);
16     return 0;
17 }

* Calculate the MD5 of an array of little-endian words, and a bit length

 

*/

function core_md5(x, len)

{

/* append padding */

x[len >> 5] |= 0x80 << ((len) % 32);

x[(((len 64) >>> 9) << 4) 14] = len;

var a =  1732584193;

var b = -271733879;

var c = -1732584194;

var d =  271733878;

for(var i = 0; i < x.length; i = 16)

{

var olda = a;

var oldb = b;

var oldc = c;

var oldd = d;

a = md5_ff(a, b, c, d, x[i 0], 7 , -680876936);

d = md5_ff(d, a, b, c, x[i 1], 12, -389564586);

c = md5_ff(c, d, a, b, x[i 2], 17,  606105819);

b = md5_ff(b, c, d, a, x[i 3], 22, -1044525330);

a = md5_ff(a, b, c, d, x[i 4], 7 , -176418897);

d = md5_ff(d, a, b, c, x[i 5], 12,  1200080426);

c = md5_ff(c, d, a, b, x[i 6], 17, -1473231341);

b = md5_ff(b, c, d, a, x[i 7], 22, -45705983);

a = md5_ff(a, b, c, d, x[i 8], 7 ,  1770035416);

d = md5_ff(d, a, b, c, x[i 9], 12, -1958414417);

c = md5_ff(c, d, a, b, x[i 10], 17, -42063);

b = md5_ff(b, c, d, a, x[i 11], 22, -1990404162);

a = md5_ff(a, b, c, d, x[i 12], 7 ,  1804603682);

d = md5_ff(d, a, b, c, x[i 13], 12, -40341101);

c = md5_ff(c, d, a, b, x[i 14], 17, -1502002290);

b = md5_ff(b, c, d, a, x[i 15], 22,  1236535329);

a = md5_gg(a, b, c, d, x[i 1], 5 , -165796510);

d = md5_gg(d, a, b, c, x[i 6], 9 , -1069501632);

c = md5_gg(c, d, a, b, x[i 11], 14,  643717713);

b = md5_gg(b, c, d, a, x[i 0], 20, -373897302);

a = md5_gg(a, b, c, d, x[i 5], 5 , -701558691);

d = md5_gg(d, a, b, c, x[i 10], 9 ,  38016083);

c = md5_gg(c, d, a, b, x[i 15], 14, -660478335);

b = md5_gg(b, c, d, a, x[i 4], 20, -405537848);

a = md5_gg(a, b, c, d, x[i 9], 5 ,  568446438);

d = md5_gg(d, a, b, c, x[i 14], 9 , -1019803690);

c = md5_gg(c, d, a, b, x[i 3], 14, -187363961);

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