多项式

/*
(x+1)的N次方的多项式
*/

#include <stdio.h>
#include <stdlib.h>
#include "POLY.h"
main(int argc, char *argv[])
  { int N = atoi(argv[1]); float p = atof(argv[2]);
    Poly t, x; int i, j;
    printf("Binomial coefficients\n");
    t = POLYadd(POLYterm(1, 1), POLYterm(1, 0));
    for (i = 0, x = t; i < N; i++)
      { x = POLYmult(t, x); showPOLY(x); }
    printf("%f\n", POLYeval(x, p));
 }
-----
typedef struct poly *Poly;
 void showPOLY(Poly);
 Poly POLYterm(int, int);
 Poly POLYadd(Poly, Poly);
 Poly POLYmult(Poly, Poly);
float POLYeval(Poly, float);
-----
#include <stdlib.h>
#include "POLY.h"
struct poly { int N; int *a; };
Poly POLYterm(int coeff, int exp)
  { int i; Poly t = malloc(sizeof *t);
    t->a = malloc((exp+1)*sizeof(int));
    t->N = exp+1; t->a[exp] = coeff;
    for (i = 0; i < exp; i++) t->a[i] = 0;    
    return t;
  }    
Poly POLYadd(Poly p, Poly q)
  { int i; Poly t;
    if (p->N < q->N) { t = p; p = q; q = t; }
    for (i = 0; i < q->N; i++) p->a[i] += q->a[i];
    return p;
  }
Poly POLYmult(Poly p, Poly q)
  { int i, j;
    Poly t = POLYterm(0, (p->N-1)+(q->N-1));
    for (i = 0; i < p->N; i++) 
      for (j = 0; j < q->N; j++) 
        t->a[i+j] += p->a[i]*q->a[j];
    return t;
  }
float POLYeval(Poly p, float x)
  { int i; double t = 0.0;
    for (i = p->N-1; i >= 0; i--) 
      t = t*x + p->a[i];
    return t;
  }

缺少 showPOLY() 的实现，话说也没弄明白这个多项式是怎么计算的。
说是计算 (x+1) 的 N 次方，但是怎么和手算出来的不一样 - -!