【c++】PAT (Advanced Level)1009. Product of Polynomials (25)

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

32000 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 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

1.写公式的时候太理所当然了，b[i+m]=n*a[i]+b[i+m]; 忘记加上原来的数字了

2.数组溢出， 1000次方和1000次方是两千次方，所以b[i]溢出。

#include <iostream>#include <iomanip>using namespace std;int main() {int k;cin>>k;double a[1001]={0.0},b[2001]={0.0};while(k--){int m;double n;cin>>m>>n;a[m]=n;}cin>>k;int i;while(k--){int m;double n;cin>>m>>n;for( i=0;i<=1000;i++){if(a[i]!=0){b[i+m]=n*a[i]+b[i+m];   }}}k=0;for( i=0;i<=2000;i++){if(b[i]!=0){k++;}}cout<<k;for(i=2000;i>=0;i--){if(b[i]!=0){cout<<" "<<i<<" "<<fixed<<setprecision(1)<<b[i];}}system("pause");return 0;}