1009. Product of Polynomials (25)

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1009. Product of Polynomials (25)

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

注意负数的情况！

#include <bits/stdc++.h>
using namespace std;
#define inf 0x7ffffff
#define ms(x,y) memset(x,y,sizeof(x))
const int M=1e6;
int i,j,n,m;
long long  sum;
struct node
{
    int no;
    double data;
};
node a[M],b[M];
double ans[M];

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        int c;double d;
        scanf("%d %lf",&c,&d);
        a[i].no=c;a[i].data=d;
    }
    scanf("%d",&m);
    for(int i=0;i<m;i++){
        int c;double d;
        scanf("%d %lf",&c,&d);
        b[i].no=c;b[i].data=d;
    }
    int flag=0;
    ms(ans,0);
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            int c;double d;
            c=a[i].no+b[j].no;
            d=a[i].data*b[j].data;
            ans[c]+=d;
            if(c>flag)flag=c;
        }
    }
    int sum=0;
    for(int i=flag+1;i>=0;i--){
        if(ans[i]!=0)sum++;
     }
      printf("%d",sum);
      for(int i=flag;i>=0;i--){
        if(ans[i]!=0)printf(" %d %.1lf",i,ans[i]);
      }
     printf("\n");
    return 0;
}