POJ1060 Modular multiplication of polynomials解题报告 (2011-12-09 20:27:53)

Modular multiplication of polynomials

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3239		Accepted: 1459

Description

Consider polynomials whose coefficients are 0 and 1. Addition of two polynomials is achieved by 'adding' the coefficients for the corresponding powers in the polynomials. The addition of coefficients is performed by addition modulo 2, i.e., (0 + 0) mod 2 =
0, (0 + 1) mod 2 = 1, (1 + 0) mod 2 = 1, and (1 + 1) mod 2 = 0. Hence, it is the same as the exclusive-or operation.

(x^6 + x^4 + x^2 + x + 1) + (x^7 + x + 1) = x^7 + x^6 + x^4 + x^2

Subtraction of two polynomials is done similarly. Since subtraction of coefficients is performed by subtraction modulo 2 which is also the exclusive-or operation, subtraction of polynomials is identical to addition of polynomials.

(x^6 + x^4 + x^2 + x + 1) - (x^7 + x + 1) = x^7 + x^6 + x^4 + x^2

Multiplication of two polynomials is done in the usual way (of course, addition of coefficients is performed by addition modulo 2).

(x^6 + x^4 + x^2 + x + 1) (x^7 + x + 1) = x^13 + x^11 + x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1

Multiplication of two polynomials f(x) and g(x) modulo a polynomial h(x) is the remainder of f(x)g(x) divided by h(x).

(x^6 + x^4 + x^2 + x + 1) (x^7 + x + 1) modulo (x^8 + x^4 + x^3 + x + 1) = x^7 + x^6 + 1

The largest exponent of a polynomial is called its degree. For example, the degree of x^7 + x^6 + 1 is 7.

Given three polynomials f(x), g(x), and h(x), you are to write a program that computes f(x)g(x) modulo h(x).

We assume that the degrees of both f(x) and g(x) are less than the degree of h(x). The degree of a polynomial is less than 1000.

Since coefficients of a polynomial are 0 or 1, a polynomial can be represented by d+1 and a bit string of length d+1, where d is the degree of the polynomial and the bit string represents the coefficients of the polynomial. For example, x^7 + x^6 + 1 can be
represented by 8 1 1 0 0 0 0 0 1.

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of three lines that contain three polynomials f(x), g(x), and h(x), one per line. Each polynomial is represented as described
above.

Output

The output should contain the polynomial f(x)g(x) modulo h(x), one per line.

Sample Input

2 7 1 0 1 0 1 1 1 8 1 0 0 0 0 0 1 1 9 1 0 0 0 1 1 0 1 1 10 1 1 0 1 0 0 1 0 0 1 12 1 1 0 1 0 0 1 1 0 0 1 0 15 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1

Sample Output

8 1 1 0 0 0 0 0 1 14 1 1 0 1 1 0 0 1 1 1 0 1 0 0

题目是说以0 1给3个数，

先按照它的乘法规则把头两个数相乘，再把结果除以第三个数求余。

对于它的加法，就是对应为上0+0＝0，0+1＝1，1+0＝1，1+1＝0.

这我们可以用位运算的异或运算符“^”来完成。

减法和加法是相同的……

乘法就是说x^6 X x^7结果x^13.按照一般乘法步骤是先两个数每个位上的数字相乘后加到对应的位置上。

求余我们可以用减法代替。

像例子中的(x^6 + x^4 + x^2 + x + 1) (x^7 + x + 1) modulo (x^8 + x^4 + x^3 + x + 1) = x^7 + x^6 + 1

其中(x^6 + x^4 + x^2 + x + 1) (x^7 + x + 1)=x^13 + x^11 + x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1

x^13 + x^11 + x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1 对应序列是1 0 1 0 1 1 0 1 1 1 1 0 0 1

x^8 + x^4 + x^3 + x + 1 对应序列是1 0 0 0 1 1 0 1 1.下面是求余过程：

1) 1 0 1 0 1 1 0 1 1 1 1 0 0 1
- 1 0 0 0 1 1 0 1 1（减）

－－－－－－－－－－－－－

...0 0 1 0 0 0 0 0 0（结果）

....2) 1 0 0 0 0 0 0 1 1（后取两位补上）

....- 1 0 0 0 1 1 0 1 1（减）

.....－－－－－－－－－－－－－

.......0 0 0 0 1 1 0 0 0 （结果）

...............1 1 0 0 0 0 0 1（补上位后发现不够长度，所以算完了）

上述所说的补位，其实可以用个变量j指定被减的开始位置，当位置小于某个数时停止。

最后要注意最后余数可能是0，所以要对此加以判断。

还有数组要开到2000……因为两个1000长度的数相乘最长是2000.

#include <iostream>
using namespace std;

int main()
{char f[4][2046];
 int n,i,m,l[3],j;
 cin>>n;
 while (n--)
 {memset(f,0,sizeof(f));
  for (i=0;i<3;i++)
  {cin>>l[i];
   l[i]-=1;
  for (m=l[i];m>=0;m--) {cin>>f[i][m];f[i][m]-=48;}
  }
  for (i=l[0];i>=0;i--)
   for (m=l[1];m>=0;m--)
   f[3][i+m]=f[3][i+m]^(f[0][i]&&f[1][m]);//mutiply
   j=l[0]+l[1];
  while (f[3][j]==0) if (j) j--;else j=-1;
  while (j>=l[2])
  {for (i=j;i>=j-l[2];i--) 
   f[3][i]^=f[2][i-j+l[2]];
   while (f[3][j]==0) if (j) j--;else j=-1;
  }
  if (j>=0)
  {cout<<j+1;
  for (i=j;i>=0;i--) cout<<' '<<f[3][i]+0;}
  else cout<<"1 0";
  cout<<endl;
 }
 return 0;
}