[swustoj 679] Secret Code

Secret Code

问题描述

The Sarcophagus itself is locked by a secret numerical code. When somebody wants to open it, he must know the code and set it exactly on the top of the Sarcophagus. A very intricate mechanism then opens the cover. If an incorrect code is entered, the tickets inside would catch fire immediately and they would have been lost forever. The code (consisting of up to 100 integers) was hidden in the Alexandrian Library but unfortunately, as you probably know, the library burned down completely.

But an almost unknown archaeologist has obtained a copy of the code something during the 18th century. He was afraid that the code could get to the ``wrong people'' so he has encoded the numbers in a very special way. He took a random complex number B that was greater (in absolute value) than any of the encoded numbers. Then he counted the numbers as the digits of the system with basis B. That means the sequence of numbers an, an-1, ..., a1, a0 was encoded as the number

X = a0 + a1B + a2B2 + ...+ anBn.

Your goal is to decrypt the secret code, i.e. to express a given number X in the number system to the base B. In other words, given the numbers X and Byou are to determine the ''digit'' a0 throughan.

输入

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case consists of one single line containing four integer numbers Xr, Xi, Br, Bi (|Xr|,|Xi| <= 1000000, |Br|,|Bi| <= 16). These numbers indicate the real and complex components of numbers X and B, i.e. X = Xr + i.Xi, B = Br + i.Bi. B is the basis of the system (|B| > 1), X is the number you have to express.

输出

Your program must output a single line for each test case. The line should contain the ''digits'' an, an-1, ..., a1, a0, separated by commas. The following conditions must be satisfied:

• for all i in {0, 1, 2, ...n}: 0 <= ai < |B|

• X = a0 + a1B + a2B2 + ...+ anBn

• if n > 0 then an <> 0

• n <= 100

If there are no numbers meeting these criteria, output the sentence "The code cannot be decrypted.". If there are more possibilities, print any of them.

样例输入

4
-935 2475 -11 -15
1 0 -3 -2
93 16 3 2
191 -192 11 -12

样例输出

8,11,18
1
The code cannot be decrypted.
16,15

同HDU 1111

题意：给定复数s和复数k，求整数序列ai使得\(s = a_{0}*k^0 + a_{1}*k^1 + a_{2}*k^2 + ...+ a_{n}*k^n\)，其中\(n<=100\),\(0<=a_{i}<|k|\),\(|k|>1\)
分析：整式变型得：\(s=a_{0}+(a_{1}+(a_{2}+...)*k)*k\)

复习一下复数运算：
A、若\(z=a+b*i\)
     则\(|z|=\sqrt{a^{2}+b^{2}}\)
B、若\(z1=a+b*i\),\(z2=c+d*i\)
     则：\(z1/z2\)
          \(=(a+b*i)/(c+d*i)\)
          \(=(a+b*i)*(c-d*i)/[(c+d*i)*(c-d*i)]\) 同时乘分母的共轭复数
          \(=(ac+bd)/(c^2+d^2)+[(bc-ad)/(c^2+d^2)]*i\)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define N 10010

struct Complex
{
    ll x,y;
    Complex(){}
    Complex(ll x,ll y):x(x),y(y){}
    ll mode2(){return x*x+y*y;}
    Complex operator - (ll t){
        return Complex(x-t,y);
    }
    bool operator % (Complex t){
        ll t1=(x*t.x+y*t.y)%t.mode2();
        ll t2=(y*t.x-x*t.y)%t.mode2();
        return t1||t2;
    }
    Complex operator / (Complex t){
        return Complex((x*t.x+y*t.y)/t.mode2(),(y*t.x-x*t.y)/t.mode2());
    }
}s,k;
ll UP;
ll flag;
ll ans[N],ansd;

void dfs(Complex now,ll step)
{
    if(flag) return;
    if(step>) return;
    if(!now.mode2())
    {
        flag=;
        ansd=step;
        return;
    }
    for(ll i=;i*i<UP && !flag;i++)
    {
        Complex tmp=now-i;
        if(tmp%k==)
        {
            ans[step]=i;
            dfs(tmp/k,step+);
        }
    }
}
int main()
{
    ll T;
    scanf("%lld",&T);
    while(T--)
    {
        flag=;
        scanf("%lld%lld%lld%lld",&s.x,&s.y,&k.x,&k.y);
        UP=k.mode2();
        dfs(s,);
        if(!flag)
            printf("The code cannot be decrypted.\n");
        else
        {
            if(!ansd) printf(""); //一开始就为0
            for(ll i=ansd-;i>=;i--)
            {
                if(i!=ansd-) printf(",");
                printf("%lld",ans[i]);
            }
            printf("\n");
        }
    }
    return ;
}