时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output
3 3 3.6 2 6.0 1 1.6
#include
#include
#include
#define MAXEXPONEN 1000
int main()
{
double A[MAXEXPONEN+1];
double B[MAXEXPONEN+1];
double result[MAXEXPONEN*2+1];
int An,Bn,e,i,j;
double c;
while(scanf("%d",&An)!=EOF){
memset(A,0,sizeof(A));
memset(B,0,sizeof(A));
memset(result,0,sizeof(A));
for(i=0;i<An;i++){
scanf("%d",&e);
scanf("%lf",&A[e]);
}
scanf("%d",&Bn);
for(j=0;j<Bn;j++){
scanf("%d",&e);
scanf("%lf",&B[e]);
}
for(i=0;i<=1000;i++){
for(j=0;j<=1000;j++){
result[i+j] += A[i]*B[j];
}
}
int count=0;
for(i=0;i<=2000;i++)
if(result[i]!=0)
count ++;
printf("%d",count);
for(i=2000;i>=0;i--){
if(result[i]!=0){
printf(" %d %.1lf",i,result[i]);
}
}
printf("\n");
}
return 0;
}