PAT (Advanced Level)1009解
因为最近在刷题库,想想就把本人可以想到的解法写到博客里,作为整理归纳。未必是最优解,还请各位高手多多包涵,能够指点指点。
题目要求
1009 Product of Polynomials (25 分)
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:
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
解题思路
按照多项式笔算写法计算,题意如图
注意事项
采用数组暴力解法的时候,需要主要到多项式相乘,原来指数最大值为1000,相乘后最大值为2000,所以上限要跟着改变。
代码部分
#include<cstdio>
main(){
float A[1001]={0.0},B[1001]={0.0},C[2002]={0.0};
//A数组用来存放第一个,B数组用来存放第二个,C数组用来存放结果
int ka,kb;
scanf("%d",&ka);
int x;
double y;
for(int i=0;i<ka;i++){
scanf("%d",&x);
scanf("%lf",&y);
A[x]=y;
}
scanf("%d",&kb);
for(int i=0;i<kb;i++){
scanf("%d",&x);
scanf("%lf",&y);
B[x]=y;
}
for(int i =0;i<=1000;i++){
for(int j=0;j<=1000;j++){
C[i+j]+=A[i]*B[j];
}
}
x=0;
for(int i =0;i<=2001;i++){
if(C[i]!=0.0)
x++;
}
printf("%d",x);
for(int i =2001;i>=0;i--){
if(C[i]!=0.0){
printf(" %d %.1lf",i,C[i]);
}
}
}
运行结果