A1002. A+B for Polynomials

题目描述#

　　This time, you are supposed to find A+B where A and B are two polynomials

输入格式#

　　Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 ... NK a_NK , where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1,2,...,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_k<...<N₂<N₁ ≤1000

输出格式#

　　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.

输入样例#

2 1 2.4 0 3.2

2 2 1.5 1 0.5

输出样例#

3 2 1.5 1 2.9 0 3.2

题意#

• 给出两行，每行表示一个多项式，第一个数表示该多项式中系数非零项的项数，后面每两个数表示一项，这两个数分别表示该项的幂次和系数。试求两个多项式的和，并以与前面相同的格式输出相同结果

样例解释#

• 是第一个2是指多项式有两个系数非零项，1 2.4 表示一次项的系数为2.4，0 3.2 表示零次项的系数（即常数项）为3.2，即有F1(x)=2.4x + 3.2
• 同样可以写出第二行所表示的多项式 F2(x)=1.5x² + 0.5x
• 那么 F1(x) + F2(x)=1.5x² + 2.9x + 3.2 

#include <bits/stdc++.h>
const int max_n = 1111;
double p[max_n] = {};
int main(int argc, char *argv[]) {
    int k, n, count = 0;
    double a;
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        scanf("%d %1f", &n, &a);
        p[n] += a;
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        scanf("%d %1f", &n, &a);
        p[n] += a;
    }
    for(int i = 0; i < max_n; i++){
        if(p[i] != 0){
            count++;
        }
    }
    printf("%d", count);
    for(int i = max_n - 1; i >= 0; i--){
        if(p[i] != 0){
            printf( "%d %.1f", i, p[i]);
        }
    }
    return 0;    
}