PAT 解题报告 1009. Product of Polynomials (25)

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

题目描述：

给定两个多项式的系数和指数, 算着两个多项式的乘积.

算法分析：

思路1：map保存项

用map保存项，计算结果需要剔除map项值==0的项。

思路2：hash保存项

注意点：

用map保存项，结果需要剔除map项值==0的项。

map

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>

using namespace std;

int main()
{
int K, L;
map<int,double> m1,m2,mp;
scanf("%d", &K);
for (int i=0; i<K; i++){
int e;double c;
scanf("%d%lf", &e, &c);
m1[e] = c;
}
scanf("%d", &L);
for (int i=0; i<L; i++) {
int e;double c;
scanf("%d%lf", &e, &c);
m2[e] = c;
}

map<int,double>::iterator it1,it2,it;
for (it1=m1.begin(); it1!=m1.end(); it1++) {
for (it2=m2.begin(); it2!=m2.end(); it2++) {
int key = it1->first + it2->first;
double value = it1->second * it2->second;
it = mp.find(key);
if (it == mp.end()) {
mp[key] = value;
}
else {
it->second += value;
}
}
}
for(it=mp.begin(); it!=mp.end();) {
if (it->second == 0.0) {
mp.erase(it++);
}
else it++;
}
printf("%d", mp.size());
map<int,double>::reverse_iterator rit;
for (rit=mp.rbegin(); rit!=mp.rend(); rit++) {
printf(" %d %.1lf", rit->first, rit->second);
}
return 0;
}

hash

#include<cstdio>
#include<cstring>

#define MAXN 2001
double hash[MAXN],hash1[MAXN];

int main(){
freopen("in.txt","r",stdin);
int n1,n2,i,j,k;
double tmp;
//scanf("%d",&n1);
while(scanf("%d",&n1)!=EOF){
memset(hash,0,sizeof(hash));
memset(hash1,0,sizeof(hash1));
for(i=0;i<n1;i++){
scanf("%d%lf",&k,&tmp);
hash1[k]=tmp;
}
scanf("%d",&n2);
for(i=0;i<n2;i++){
scanf("%d%lf",&k,&tmp);
for(j=1000;j>=0;j--){
hash[j+k]+=hash1[j]*tmp;
}
}
int count=0;
for(i=0;i<MAXN;i++){
if(hash[i]!=0) count++;
}
if(count==0) printf("0\n");
else{
printf("%d",count);
for(i=MAXN-1;i>=0;i--){
if(hash[i]!=0) printf(" %d %.1lf",i,hash[i]);
}
printf("\n");
}
}
return 0;
}