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

分析：

简单模拟～double类型的arr数组保存第一组数据，ans数组保存结果。当输入第二组数据的时候，一边进行运算一边保存结果。最后按照指数递减的顺序输出所有不为0的项～

注意：

因为相乘后指数可能最大为2000，所以ans数组最大要开到2001～

原文链接：https://blog.csdn.net/liuchuo/article/details/52109341

题解

找了半天bug，发现ans数组开了2001，，，，，
因为ans需要把两个数组的下标相加，所以要注意这点。。。

#include <bits/stdc++.h>

using namespace std;
double b,arr[1001],ans[2001];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n1,n2,a,cnt=0;
    scanf("%d",&n1);
    for(int i=0;i<n1;i++){
        scanf("%d %lf",&a,&b);
        arr[a]=b;
    }
    scanf("%d",&n2);
    for(int i=0;i<n2;i++){
        scanf("%d %lf",&a,&b);
        for(int j=0;j<1001;j++){
            ans[j+a]+=arr[j]*b;
        }
    }
    for(int i=2000;i>=0;i--)
        if(ans[i]!=0) cnt++;
    printf("%d",cnt);
    for(int i=2000;i>=0;i--){
        if(ans[i]!=0) printf(" %d %.1f",i,ans[i]);
    }
    return 0;
}