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 //多项式的乘法
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAX=1010;
int g[MAX],f[MAX],h[MAX],gf[MAX*MAX];//gf[@@] 空间一定要开大，否则aw;
int a,b,c;
int d;
//写一个比较函数
int compare(){
if(d<c)
return -1;
if(d>c)
return 1;
for(int i=d-1;i>=0;i--){
if(gf[i]==h[i])
continue;
else if(gf[i]>h[i])
return 1;
else if(gf[i]<h[i])
return -1;
}
return 0;//两者完全相同
}
int main(){
int t;
cin>>t;
while(t--){
memset(h,0,sizeof(h));
memset(gf,0,sizeof(gf));//不能少，否则会第二个实例结果会出现问题
cin>>a;
a--;
for(int i=a;i>=0;i--){
cin>>g[i];//存入g(x)函数
}
cin>>b;
b--;
for(int i=b;i>=0;i--){
cin>>f[i];//存入f(x)函数
}
cin>>c;
c--;
for(int i=c;i>=0;i--){
cin>>h[i];//存入h(x)函数
}
//计算乘法
for(int i=a;i>=0;i--)
for(int j=b;j>=0;j--){
gf[i+j]=gf[i+j]^(g[i]&f[j]);
}
d=a+b;
//求余就是多项式相减到小于除数
while(compare()>=0){
int e=d-c;//长度相减
for(int i=d; i-e>=0; i--){
gf[i]=gf[i]^h[i-e];
}
while(d&&(gf[d]==0))
d--;
}
cout<<d+1;
for(int i=d;i>=0;i--)
cout<<" "<<gf[i];
cout<<endl;
}
}
/*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
*/