This time, you are supposed to find A*B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=1000.
Output Specification:
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
Sample Input
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1 2 |
2 1 2.4 0 3.2 2 2 1.5 1 0.5 |
Sample Output
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1 |
3 3 3.6 2 6.0 1 1.6 |
代码如下：
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#include<stdio.h> #include<stdlib.h> double result[2002]; typedef struct { int exponent; double coefficient; } Polynomial; int main() { int i,j,n1,n2,count,index; Polynomial *a,*b; double temp,sum; scanf("%d",&n1); a=(Polynomial *)malloc(n1*sizeof(Polynomial)); for(i=0;i<n1;i++) { scanf("%d%lf",&a[i].exponent,&a[i].coefficient); } scanf("%d",&n2); b=(Polynomial *)malloc(n2*sizeof(Polynomial)); for(i=0;i<n2;i++) { scanf("%d%lf",&b[i].exponent,&b[i].coefficient); } for(i=0;i<n1;i++) { for(j=0;j<n2;j++) { index=a[i].exponent+b[j].exponent; temp=a[i].coefficient*b[j].coefficient; result[index]+=temp; } } for(i=2001;i>=0;i--) { if(result[i]!=0) { count++; index=i; } } if(count==0) printf("0\n"); /* else if(count==1) printf("1 %d %.1lf\n",index,result[index]);*/ else { printf("%d ",count); for(i=2001;i>index;i--) { if(result[i]!=0) printf("%d %.1lf ",i,result[i]); } printf("%d %.1lf",i,result[i]); } free(a); free(b); } |