【PAT-甲级-C++】1002. A+B for Polynomials (25)

1002. A+B for 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

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

2、注意

3、代码

#include
#define N 1001
using namespace std;
int main() {
    int n1, n2,i,j;
    float f[N] = { 0 };
    int a;
    float b;
    int max = 0, num = 0;
    cin >> n1;
    for (i = 0;i < n1;++i) {
        scanf("%d %f", &a, &b);
        f[a] += b;
        if (a > max)
            max = a;
    }
    cin >> n2;
    //读入数据
    for (i = 0;i < n2;++i) {
        scanf("%d %f", &a, &b);
        f[a] += b;
        if (a > max)
            max = a;
    }
    //求解非零总项数
    for (i = max;i >= 0;--i) {
        if (f[i] != 0) {
            num++;
        }
    }
    //输出结果
    printf("%d", num);
    for (i = max;i >= 0;--i) {
        if (f[i] != 0) {
            printf(" %d %.1f", i,f[i]);
        }
    }
    system("pause");
    return 0;
}