1 A+B for Polynomials

题目链接：1002 A+B for Polynomials

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:
$K\ N_1\ a_{​N​1}​​​​ N_2\ a_{​N​2}​​​​...\ N_K​\ a_{​N​K}​​​​$
where $K$ is the number of nonzero terms in the polynomial, $N​_i$​​ and $a​_{N​i}​​​​(i=1,2,⋯,K)$ are the exponents and coefficients, respectively. It is given that
$1 ≤ K ≤ 10，0 ≤ N​_K ​​< ⋯ < N_2 ​​< N​_1 ≤ 1000$.

Output Specification

For each test case you should output the sum 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 to 1 decimal place.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

解题思路

这道题有几个点需要注意：

• 处理输入数据方式和常规有一些区别
• coefficients 为浮点数且为负数时也要输出
• 输出要精确到小数点后一位
• 最后没有空格

题解

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 1e3 + 5;
double a[maxn] = {0};

int main(){
    #ifdef ONLINE_JUDGE
    #else
        freopen("input.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int k;
    cin >> k;
    for(int i = 0;i < k;i++){
        int exp;
        double coe;
        cin >> exp >> coe;
        a[exp] = coe;
    }
    cin >> k;
    for(int i = 0;i < k;i++){
        int exp;
        double coe;
        cin >> exp >> coe;
        a[exp] += coe;
    }
    int cnt = 0;
    for(int i = 0;i < maxn;i++){
        if(a[i]){
            cnt++;
        }
    }
    printf("%d", cnt);
    for(int i = maxn;i >= 0;i--){
        //负数也需要输出
        if(a[i] != 0){
			//cout << " " << i << " " << a[i];
            //coe精确到小数点后一位
            printf(" %d %.1f", i, a[i]);
        }
    }
    return 0;
}

2 Product of Polynomials

题目链接：1009 Product of Polynomials

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:
$K\ N_1\ a_{​N​1}​​​​ N_2\ a_{​N​2}​​​​...\ N_K​\ a_{​N​K}​​​​$
where $K$ is the number of nonzero terms in the polynomial, $N​_i$​​ and $a​_{N​i}​​​​(i=1,2,⋯,K)$ are the exponents and coefficients, respectively. It is given that
$1 ≤ K ≤ 10，0 ≤ N​_K​ ​< ⋯ < N_2 ​​< N​_1 ≤ 1000$.

Output Specification

For each test case you should output the sum 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 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

注意一个坑：作为输出结果的 $C[maxn2]$ 数组，范围需要开到 $2000$ 以上才可。

题解

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 1005, maxn2 = 2010;
double a[maxn] = {0}, b[maxn] = {0}, c[maxn2] = {0};

int main(){
    int k;
    cin >> k;
    for(int i = 0;i < k;i++){
        int exp;
        double coe;
        cin >> exp >> coe;
        a[exp] = coe;
    }
    cin >> k;
    for(int i = 0;i < k;i++){
        int exp;
        double coe;
        cin >> exp >> coe;
        b[exp] = coe;
    }
    for(int i = 0;i < maxn;i++){
        for(int j = 0;j < maxn;j++){
            c[i + j] += a[i] * b[j];
        }
    }
    int cnt = 0;
    for(int i = 0;i < maxn2;i++){
        if(c[i]){
            cnt++;
        }
    }
    cout << cnt;
    for(int i = maxn2;i >= 0;i--){
        if(c[i] != 0){
            printf(" %d %.1f", i, c[i]);
        }
    }
    return 0;
}