POJ 1060 Modular multiplication of polynomials

Method (no proof, LZ weak slag please understand):

In the example: (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.

(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.

Make a=x^13 + x^11 + x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1, b=x^8 + x^4 + x^3 + x + 1;

For the remainder of A/b, first b*x^ (13-8), since the coefficients are 0 or 1, so a= b* (x^ (13-8)) XOR a=x^11+x^4+x^3+1

11>8 Repeat the steps above: a= b* (x^ (11-8)) XOR a=x^7+x^6+x^5+1

7<8 structure, therefore (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= x^7+x^6+x^5+1

This is not the main thing to do, the others are good to write

#include <iostream>#include<cstdio>#include<cstring>using namespacestd;Const intmaxn= .;//F,g,h stores the coefficients of the polynomial, sum stores the coefficients of the f*g and the coefficients of the last remainder.intF[MAXN],G[MAXN],H[MAXN],SUM[MAXN];intLf,lg,lh,ls;//the highest power of the F,g,h,sum, respectively.//Compare the size of the polynomial represented by the sum with the H representation of the polynomialintCompare () {if(ls<LH)return-1; if(ls>LH)return 1;  for(inti=ls-1; i>=0; i--) {        if(sum[i]==H[i])Continue; if(sum[i]>H[i])return 1; if(sum[i]<H[i])return-1; }    return 0;}intMain () {intt,d; scanf ("%d",&t);  while(t--) {memset (H,0,sizeof(h)); memset (SUM,0,sizeof(sum)); //depositing the information of the F-polynomial into the F-arrayscanf"%d",&d); LF=d-1;  for(intJ=LF; j>=0; j--) {scanf ("%d",&F[j]); }        //The G-polynomial information is stored in the G arrayscanf"%d",&d); LG=d-1;  for(intJ=LG; j>=0; j--) {scanf ("%d",&G[j]); }        //The H -polynomial information is stored in the H arrayscanf"%d",&d); LH=d-1;  for(intJ=LH; j>=0; j--) {scanf ("%d",&H[j]); }        //polynomial for calculating f*gls=lf+LG;  for(intI=LF; i>=0; i--) {             for(intJ=LG; j>=0; j--) {Sum[i+j]=sum[i+j]^ (f[i]&G[j]); }        }        /*The key is how to find the remainder, here is the first to determine whether the sum polynomial is greater than the H polynomial, if greater than, then sum minus one time H, minus the information into sum.          Continue judging until sum is less than H, then the sum is the remainder.        In short, it is to change the division to a simpler subtraction operation. */         while(Compare () >=0) {D=ls-LH;  for(intI=ls; i-d>=0; i--) {Sum[i]=sum[i]^h[i-d]; }             while(LS &&!)sum[ls]) LS--; } printf ("%d", ls+1);  for(intI=ls; i>=0; i--) {printf ("%d", Sum[i]); } printf ("\ n"); }    return 0;}

Reference http://www.cnblogs.com/chenxiwenruo/p/3332698.html

POJ 1060 Modular multiplication of polynomials