A1002 A+B for Polynomials (25)（25 分）

1002 A+B for Polynomials (25)（25 分）

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 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 <= 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

思考

这里胡凡书上写的不全面，代码修正如下。

AC代码

#include <stdio.h>
#define max_n 1111
double p[max_n] = {};//这样也可以置初值为0 
int main() {
    int k, n, count = 0;
    double a;
    scanf("%d", &k);
    count +=k; 
    for(int i = 0; i < k; i++) {
        scanf("%d %lf", &n, &a);
        p[n] += a;
    }
    scanf("%d", &k);
    count +=k; 
    for(int i = 0; i < k; i++) {
        scanf("%d %lf", &n, &a);
        if(p[n]!=0) count--;//出现重合项-1
        p[n] += a;
        if(p[n]==0) count--;//出现重合项且一正一负抵消为0，再-1
    }
//    for(int i = 0; i < max_n; i++) {
//        if(p[i] != 0) {
//            count++;
//        }//这样计数非零项是保险的 
//    }
    printf("%d", count);
    for(int i = max_n - 1; i >= 0; i--) {
        if(p[i] != 0) printf(" %d %.1f", i, p[i]);
    }
    return 0;
}