PAT甲级 1009

Product of Polynomials

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

---

当保留一位小数时，注意判断是否为零时要使用fabs(double).

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
const int maxn=2000+2;
double inputA[maxn>>1],inputB[maxn>>1],result[maxn];
int main(){
    //freopen("./in","r",stdin);
    int k;
    scanf("%d",&k);
    int i,j;
    int n;
    for(i=0;i<k;i++){
        scanf("%d",&n);
        scanf("%lf",inputA+n);
    }
    scanf("%d",&k);
    for(i=0;i<k;i++){
        scanf("%d",&n);
        scanf("%lf",inputB+n);
    }
    for(i=0;i<(maxn>>1);i++){
        for(j=0;j<(maxn>>1);j++){
            result[i+j]+=inputA[i]*inputB[j];
        }
    }
    int cnt=0;
    for(i=0;i<maxn;i++){
        if(fabs(result[i])-0.5>1e-6){
            cnt++;
        }
    }
    printf("%d",cnt);
    for(i=maxn-1;i>=0;i--){
        if(fabs(result[i])-0.5>1e-6){
            printf(" %d %.1f",i,result[i]);
        }
    }
}