#P6959. [NEERC2017] Hack

    ID: 6091 Type: RemoteJudge 1000ms 128MiB Tried: 0 Accepted: 0 Difficulty: 7 Uploaded By: Tags>2017交互题Special JudgeACM_ICPC

[NEERC2017] Hack

题目描述

Heidi is analyzing a peculiar device. This device takes an a as input and computes ad(modn)a^d(mod n) using thefollowing pseudocode and some integers dd and nn stored in this device:

modPow(a, d, n) {
 r = 1;
 for (i = 0; i < 60; ++i) {
   if ((d & (1 << i)) != 0) {
     r = r * a % n;
   }
 a = a * a % n;
 }
}

Note that the pseudocode assumes arbitrary sized integers, <<<< denotes bitwise shift left, & denotes bitwise

and, and % denotes modulo.

The device does not tell Heidi the result of the computation. However, Heidi can measure how long does the computation take. She knows that only multiplication modulo nn (lines 55 and 77 in the above pseudocode) takes any measurable amount of time, all other lines can be assumed to take 00 nanoseconds.

Moreover, she knows that it takes (bits(x)+1)(bits(y)+1)(bits(x) + 1) · (bits(y) + 1) nanoseconds to multiply xx by yy modulo nn , where bits(x)bits(x) is the number of bits in the binary representation of xx without leading zeros, or more formally $\text{bits(x)} = ⌈\log_2 (x + 1)⌉.

Heidi knows the integer nn but does not know the integer dd . She wants to find dd by feeding the device different integers a as input and measuring the time the computation takes for each a .

She knows that nn and dd were chosen in the following way: first, two prime numbers pp and qq with 3030 bits in binary representation (in other words, between 229229 and 2301)230 −1) were picked independently and uniformly at random. Then the number nn was computed as n=pqn = p · q . Then the number m=φ(n)=(p1)(q1)m = φ(n) = (p−1)·(q −1)

was computed. Then dd was picked uniformly at random between 11 and m1m − 1 inclusive, such that it is coprime with mm .

Interaction Protocol

First, the testing system writes the integer nn -- the modulo used by the device. Note that nn and the hidden number dd are guaranteed to have been generated according to the procedure described above.

Your solution shall print requests of two types:

  • “? a” tells to feed a as input to the device. a must be an integer between 00 and n1n−1 inclusive. The testing system responds with the time it took the device to compute modPow(a , d , n) in nanoseconds.

  • “! d” tells the value of dd that your program has determined.

Don't forget to flush the output after each request!

Your solution must issue exactly one request of the second type, which must be the last request, and the solution must terminate gracefully after issuing it.

Your solution is allowed to issue at most 3000030 000 requests of the first type.

Your solution will be run on 3030 testcases, working with one (n,d)(n , d) pair per run. For each testcase the numbers nn and dd are fixed and were generated using the procedure described above. The example below

was not generated in that manner and thus will not be used for testing your solution; it only serves to illustrate the input/output format and provide a sanity check for your calculation of the computation time.

题目大意

有如下的一个程序来计算 ad(mod n)a^d(mod \ n),(d,nd,n是常数)

modPow(a, d, n) {
	r = 1;
 	for (i = 0; i < 60; ++i) {
   		if ((d & (1 << i)) != 0) {
     		r = r * a % n;
   		}
 		a = a * a % n;
 	}
}

其中,计算 xy  mod nx * y\ \ mod \ n (上述伪代码中的第 55 行和第 77 行)需要消耗 (bits(x)+1)(bits(y)+1)(bits(x) + 1)* (bits(y) + 1) 纳秒,bits(x)bits(x)xx 的二进制表示中不带前导零的位数,更正式的说,为 log2(x+1)\lceil \log_2(x+1) \rceil,其他指令可以认为不需要任何时间。

你知道 nn ,但不知道 dd, 你可以通过不超过 3000030000 次询问对于 aa 计算 ad(mod n)a^d(mod \ n) 所用纳秒数。

正式数据中,n,dn,d 的生成方式如下:随机挑选两个 [229,2301][2^{29},2^{30}-1] 的质数 p,qp,qn=pqn=pq,而 dd 为在 [1,φ(n)][1,\varphi(n)] 随机挑选的,与 φ(n)=(p1)(q1)\varphi(n)=(p-1)*(q-1) 互质的数

这是一道交互题

首先,将给出整数 nn

有两种指令可用:

“? a”询问对于正整数 aa 计算 ad(mod n)a^d(mod \ n) 所用纳秒数。要求a<na<n。返回所用的纳秒数

“! d”表示确定了 dd 的值并提交,你的程序应当在此后结束。

15
980
293
? 3
? 8
! 5