1002. A+B for Polynomials (25)-PAT甲级真题

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

Input

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

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

题目大意：计算多项式A+B的和～

分析：设立c数组，长度为指数的最大值，c[i] = j表示指数i的系数为j，接收a和b输入的同时将对应指数的系数加入到c中，累计c中所有非零系数的个数，然后从后往前输出所有系数不为0的指数和系数～

#include <iostream>
using namespace std;
int main() {
    float c[1001] = {0};
    int m, n, t;
    float num;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        scanf("%d%f", &t, &num);
        c[t] += num;
    }
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d%f", &t, &num);
        c[t] += num;
    }
    int cnt = 0;
    for (int i = 0; i < 1001; i++) {
        if (c[i] != 0) cnt++;
    }
    printf("%d", cnt);
    for (int i = 1000; i >= 0; i--) {
        if (c[i] != 0.0)
            printf(" %d %.1f", i, c[i]);
    }
    return 0;
}

#include <iostream>

using namespace std;

int main() {

float c[1001] = {0};

int m, n, t;

float num;

scanf("%d", &m);

for (int i = 0; i < m; i++) {

scanf("%d%f", &t, &num);

c[t] += num;

}

scanf("%d", &n);

for (int i = 0; i < n; i++) {

scanf("%d%f", &t, &num);

c[t] += num;

}

int cnt = 0;

for (int i = 0; i < 1001; i++) {

if (c[i] != 0) cnt++;

}

printf("%d", cnt);

for (int i = 1000; i >= 0; i--) {

if (c[i] != 0.0)

printf(" %d %.1f", i, c[i]);

}

return 0;

}

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