PAT 1009. Product of Polynomials (25)（多项式乘法）（待修改）

题目

1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
65536 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 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

解题思路

• 1.由于每行的输入都是由高次到低次，并且不重复的，所以可以写的更简单，还要考虑浮点数问题，但是我感觉都考虑到了，第一个测试点还是通不过！！！不知道错在哪！
• 2.又想了下，难道是有个指数的和项大于1.e*10 -8，但是每一项不大于？测试点这么叼？

代码（测试一通不过）

#include<iostream>#include<iomanip>#include<map>#include<vector>#include<algorithm>using namespace std;int exp1[1001];double coe1[1001];map<int ,double > product;bool cmp(int a,int b){    return a>b;}const double eps = 1.e-8;int main(int argc, char *argv[]){    ios_base::sync_with_stdio(false);    //输入第一组数据    int n1;    cin >> n1;    for (int i = 0; i < n1; ++i) {        cin >> exp1[i] >> coe1[i];    }    //输入第二组数据并计算乘积    int n2;    cin >> n2;    int exp2,exp3;double coe2,coe3;    vector<int> keep;    for (int i = 0; i < n2; ++i) {        cin >> exp2 >> coe2;        for (int i = 0; i < n1; ++i) {            exp3 = exp2 + exp1[i];            coe3 = coe2 * coe1[i];            if (coe3 > eps) {                if (product[exp3] ==0) {                   product[exp3] = coe3;                   keep.push_back(exp3);                }else {                    product[exp3] += coe3;                }            }        }    }    if (!keep.empty()) {        sort(keep.begin(),keep.end(),cmp);    }    cout << keep.size();    for (int i = 0; i < keep.size(); ++i) {        cout << " " << keep[i] << " " << fixed << setprecision(1) << product[keep[i]];    }    cout << endl;    return 0;}

AC代码

#include <cstdio>#include <string>#include <cstring>#include <cmath>using namespace std;const double eps = 1.e-8;double a[1024],b[1024],c[2048];int main() {int n;    for (scanf("%d",&n);n;--n) {        int x;        scanf("%d",&x);        scanf("%lf",a + x);    }    for (scanf("%d",&n);n;--n) {        int x;            scanf("%d",&x);            scanf("%lf",b + x);        }    for (int i = 0; i <= 1000; ++i) {        for (int j = 0; j <= 1000; ++j) {            c[i + j] += a[i] * b[j];        }    }    int num = 0;    for (int i = 2000; i >= 0; --i) {        if (fabs(c[i]) >= eps) {            ++num;        }    }    printf("%d",num);    for (int i = 2000; i >= 0; --i) {        if (fabs(c[i]) >= eps) {            printf(" %d %.1lf",i,c[i]);        }    }    puts("");    return 0;}