baby_step gaint_step 算法基本思想:

对于一个n个元素的循环(n很大很大) 先算出前面m步(baby_step) 然后以m为跨度(gaint_step)大跳 那么跳了n/m步以后 一定能跳到前面算出来的m步里面 这样时间复杂度就降到O(m+n/m) 空间复杂度为O(m)

对于计算a^x==b(mod n)中的x

先计算b,b*a,b*a^2,...b*a^m 然后计算1，a^m,a^2m,a^3m,... 那么经过i步 就是到了a^(i*m)的时候 发现它等于b*a^j 那么x=i*m-j

一般m定为sqrt(n)平衡时空(并且这样时间复杂度最低） 查找用hash 事实证明map是非常慢的

//更新

经过AekdyCoin盖世神牛的检验 我那个能在poj上跑的程序在hdu上先MLE 然后TLE 然后CE 然后RE 然后WA 千辛万苦 最后跳过PE 变成AC了

原因:动态链表hash跑太慢 以后要改成前向星了

//继续更新

经过AekdyCoin教导 发现这个算法当a和n不互质的时候会死 因为没有逆元 i*m-j不能随便减

于是连夜开发不互质算法如下

设某质数p在a里的指数是ap 在n里面是np 在b里面是bp

那么当x很大 ap*x必然大于np 这个时候bp必须不小于np 其逆命题也成立

同时 将n里面的p全部除掉 剩下的由于和p互质 所以左边a，b可不必除 反正最后a^x-b一定整除n

所以 先判断a和n的公共质因数里面 有没有b比n小的 若小必死 否则直接将n除的和a互质再做完破

注意到每个质数的指数肯定不超过40 那么当x大于40以上方法必然成立 当x小的时候 虽然经证明也可以化为abn互质情况 但是不如直接验证 所以不管了

171A - Mysterious numbers - 1

The easiest way to make the problem statement unusual is to omit it. This is an extremely convenient approach — you don’t have to maintain the statement in two languages or to worry that it might turn out to be ambiguous or too long or too scary. 690 people solved this problem, so evidently we can omit statements even in regular rounds :-)

As for the problem itself, it required to sum the first number and the reverse of the second number.

171B - A star

They say it’s better to see once than to hear ten times or to read a hundred times. In this problem we decided to check this and to replace the traditional textual statement with a single image. Same as in the previous problem, it did well — at least 645 participants recognized star numbers (sequence http://oeis.org/A003154 in OEIS), the numbers of balls needed to form a six-pointed start of certain size. After this one had only to code the formula — S_n = 6n(n−1) + 1.

171D - Broken checker

What does one do if the statement is unknown and the only source of information about the problem is the checker? Right — you just try all possible functions which convert 5 input values into 3 output values and see which of them fits :-)

171F - ucyhf

This problem finally has a statement! The trick is, it’s encoded. We decided to be kind to you and to use the simplest cipher possible — Caesar cipher (each letter is shifted the same number of positions in the alphabet). By a long stretch of imagination one could break the cipher by hand — observe frequent letters and short words, deduce possible values of shift and verify it against the rest of the message. A lazier one could Goog

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int n;
    while (scanf("%d",&n)==1)
    {
        printf("%d\n",6*n*n-6*n+1);
    }
    return 0;
}

le for a tool like this one and get the decoded version semi-automatically.

After cracking the statement the rest was almost easy — you had to find a prime whose reverse is a prime different from the original one (sequence http://oeis.org/A006567). 11184-th such number equals 999983, so one could do a brute-force check of all numbers in row.

171E - UNKNOWN LANGUAGE

A special contest written by me and no special programming language? Impossible! I pulled myself up and made only 25% of all problems esoteric — that’s two. There really should be three but the third interpreter refused to cooperate. Maybe next time…

What can one do if all he knows about the language is its compiler? Just run any code and see what the compiler says. In this case the compiler said “DO YOU EXPECT ME TO FIGURE THIS OUT?”, and Google should tell you immediately that the language in question is INTERCAL. The problem simplifies to figure out the dialect used and how to output “INTERCAL” in it.

In Codeforces round #96 I gave a problem 133C - Turing Tape which explained the mechanism of string output in INTERCAL and asked to write a program which converted a string into an array of numbers which would print this string. Combine this knowledge with Hello, World! example and you get the program you need. Actually, that’s what I did to write the reference solution.

PLEASE DO ,1 <- #8 DO ,1 SUB #1 <- #110 PLEASE DO ,1 SUB #2 <- #32 DO ,1 SUB #3 <- #72 DO ,1 SUB #4 <- #136 DO ,1 SUB #5 <- #88 DO ,1 SUB #6 <- #136 DO ,1 SUB #7 <- #64 DO ,1 SUB #8 <- #80 PLEASE READ OUT ,1 PLEASE GIVE UP

171C - A Piece of Cake

The second esoteric problem had a Chef program as the statement. You had only to figure out what it does and do it in any regular language. It turns out that this program reads N followed by N numbers $a_1, a_2, …, a_N$ and calculated the sum i * a_i.

171G - Mysterious numbers - 2

This one was much harder to guess but much easier to code. First two numbers were the start of a Fibonacci-like sequence, and the third one was the index of the required number in this sequence.

#include<stdio.h>
#include<string.h>
#include<math.h>
int down(int s)
{
    int t;
    t=0;
    while (s)
    {
        t=10*t+s%10;
        s/=10;
    }
    return t;
}
int main()
{
    int a,b;
    while (scanf("%d%d",&a,&b)==2)
    {
        printf("%d\n",a+down(b));
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int n;
    while (scanf("%d",&n)==1)
    {
        printf("%d\n",6*n*n-6*n+1);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int n,i,sum,a;
    while (scanf("%d",&n)==1)
    {
        sum=0;
        for (i=1;i<=n ;i++ )
        {
            scanf("%d",&a);
            sum+=i*a;
        }
        printf("%d\n",sum);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int g[6]={0,2,3,1,2,1},i;
    while (scanf("%d",&i)==1)
    {
        printf("%d\n",g[i]);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int a,b,c;
    while (scanf("%d%d%d",&a,&b,&c)==3)
    {
        c--;
        while (c--)
        {
            b=a+b;
            a=b-a;
        }
        printf("%d\n",b);
    }
    return 0;
}