POJ2527+多项式除法

模拟一遍即可。

注意一些特殊情况，见代码。

 #include<stdio.h>
 #include<stdlib.h>
 #include<math.h>
 #include<algorithm>
 #include<string.h>
 using namespace std;

 const int maxn = ;

 struct Ploy{
     int cnt;//项的数目
     int coe[ maxn ];//各项系数
     int exp[ maxn ];//各项指数
 }a;
 struct Ploy2{
     int coe1,coe2;
     int exp1,exp2;
 }b;
 struct Ploy3{
     int coe,exp;
 }tmp;

 bool Judge( int aim ){
     //bool f = false;
     for( int i=a.cnt-;i>=;i-- ){
         if( a.coe[i]!=&&a.exp[i]>=aim ){
             //f = true;
             return true ;
         }
     }
     return false;
 }

 void solve( int n,int k ){
     //ans.cnt = 0;
     while(  ){
         //if( Judge(k)==false ) break;
         int i;
         for( i=a.cnt-;i>=;i-- ){
             if( a.coe[i]!= ){
                 tmp.coe = a.coe[i];
                 tmp.exp = a.exp[i];
                 break;
             }
         }
         if( tmp.exp<k ) break;
         int delta_exp = tmp.exp-k;
         int delta_coe = tmp.coe;//商
         b.exp1 = tmp.exp,b.coe1 = tmp.coe;
         b.exp2 = delta_exp,b.coe2 = delta_coe;
         //bool f1 = false,f2 = false;
         for( int i=;i<a.cnt;i++ ){
             if( a.exp[i]==b.exp1 ){
                 a.coe[i] -= b.coe1;
                 //f1 = true;
             }
             if( a.exp[i]==b.exp2 ){
                 a.coe[i] -= b.coe2;
                 //f2 = true;
             }
         }
     }
 }

 int main(){
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int n,k;
     while( scanf("%d%d",&n,&k)== ){
         if( n==k&&k==- ) break;
         a.cnt = ;
         for( int i=;i<=n;i++ ){
             a.cnt ++;
             scanf("%d",&a.coe[i]);
             a.exp[i] = i;
         }
         if( k== ){
             puts("");
             continue;
         }
         //printf("input\n");
         bool f = false;
         for( int i=;i<a.cnt;i++ ){
             if( a.coe[i]!= ){
                 f = true;
                 break;
             }
         }
         if( f==false ) {
             puts("");
             continue;
         }//多项式全为0
         if( k>n ){
             for( int i=;i<a.cnt;i++ ){
                 if( i== ) printf("%d",a.coe[i]);
                 else printf(" %d",a.coe[i]);
             }
             printf("\n");
             continue;
         }//商为0
         //printf("solve\n");
         solve( n,k );
         f = false;
         for( int i=;i<a.cnt;i++ ){
             if( a.coe[i]!= ){
                 f = true;
                 break;
             }
         }
         if( f==false ) {
             puts("");
             continue;
         }//刚好整除
         int pos = a.cnt-;
         for( int i=a.cnt-;i>=;i-- ){
             if( a.coe[i]!= ){
                 pos = i;
                 break;
             }
         }
         for( int i=;i<=pos;i++ ){
             if( i== ) printf("%d",a.coe[i]);
             else {
                 printf(" %d",a.coe[i]);
             }
         }
         printf("\n");
     }
     return ;
 }