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

传送门

https://www.patest.cn/contests/pat-a-practise/1002

题目

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

分析

开始想的特别多,结果发现400ms的时间循环1001个数还是很简单的,直接遍历就行。

需要注意的是:
1.输入0 0要输出一个0;
2.第一次输入没有重复的指数,直接按下标赋值到数组里面即可。

源代码

//C/C++实现
#include <iostream>

using namespace std;

float num[1001];

int main(){
    int n1, n2; //项数
    float coef; //系数
    int exp; //指数
    int max_exp = -1; //最大指数
    scanf("%d", &n1);
    for(int i = 0; i < n1; ++i){
        scanf("%d %f", &exp, &coef);
        max_exp = (exp > max_exp ? exp : max_exp);
        num[exp] = coef;
    }
    scanf("%d", &n2);
    for(int i = 0; i < n2; ++i){
        scanf("%d %f", &exp, &coef);
        max_exp = (exp > max_exp ? exp : max_exp);
        num[exp] += coef;
    }
    int count = 0; //项数计算
    for(int i = 0; i <= 1000; ++i){
        if(num[i]){
            ++count;
        }
    }
    printf("%d", count);
    for(int i = max_exp; i >= 0; --i){
        if(num[i] != 0){
            printf(" %d %.1f", i, num[i]);
        }
    }
    printf("\n");
    return 0;
}

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