目錄
- [PAT Advanced Level] 1009 Product of Polynomials
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- 題目
- 題解
- 參考
[PAT Advanced Level] 1009 Product of Polynomials
題目
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 N 1 a N 1 N 2 a N 2 . . . N K a N K K N_1 a_{N1} N_2 a_{N2}...N_K a_{NK} KN1aN1N2aN2...NKaNK,where K is the number of nonzero terms in the polynomial, N i N_i Ni and a N i a_{N_i} aNi ( i = 1 , 2 , ⋯ , K ) (i=1,2,⋯,K) (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤ N K N_K NK <⋯< N 2 N_2 N2 < N 1 N_1 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.
輸入給出兩組多項式A、B,求A*B。輸入每一行,第一位為非零項的個數,後續分别為指數、系數、指數、系數。
示例
輸入:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
輸出:
3 3 3.6 2 6.0 1 1.6
題解
先以結構體形式儲存一組多項式的系數,指數由序号表示,然後輸入第二組系數與第一組進行循環相乘,并加到對應指數的系數上,最後輸出所有非零系數的項。
答案ans的系數數組至少要開到2001,因為兩個最高次幂為1000的多項式相乘,最高幂次可以達到2000。
#include <cstdio>
struct Poly
{
int exp; //指數
double cof; //系數
} poly[1001]; //第一個多項式
double ans[2001]; //結果
int main()
{
int n,m,number=0;
scanf("%d",&n); //第一個多項式中非零系數的項數
for(int i=0;i<n;i++)
{
scanf("%d %lf",&poly[i].exp,&poly[i].cof);
}
scanf("%d",&m); //第二個多項式中非零系數的項數
for(int i=0;i<m;i++)
{
int exp;
double cof;
scanf("%d %lf",&exp,&cof); //第二個多項式的指數和系數
for(int j=0;j<n;j++)
{
ans[exp+poly[j].exp]+=(cof*poly[j].cof);
}
}
for(int i=0;i<=2000;i++)
{
if(ans[i]!=0.0) number++; //累計非零系數的項數
}
printf("%d",number);
for(int i=2000;i>=0;i--)
{
if(ans[i]!=0.0) printf(" %d %.1f",i,ans[i]); //保留1位小數輸出
}
return 0;
}
參考
[算法筆記-上機訓練實戰指南-胡凡]