1002. A+B for Polynomials (25)

最新推荐文章于 2024-01-28 22:42:58 发布

xiaobai12123

最新推荐文章于 2024-01-28 22:42:58 发布

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分类专栏： PAT(A)

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本文介绍了一种解决多项式加法问题的算法实现方法，输入两个多项式，输出它们相加后的结果。该算法使用数组存储多项式的系数，并通过扫描输入来填充这些系数，最后输出相加后的多项式。

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PAT

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

即为普通的多项式系数相加问题。按指数从大到小输出.

代码

#include<cstdio>
int main(){
    double a[1024]={0};
    double t=0;
    int i=0,l=0,cnt1=0,cnt2=0;
    scanf("%d",&cnt1);
    for(i=0;i<cnt1;i++){
        scanf("%d",&l);
        scanf("%lf",&a[l]);
    }
    scanf("%d",&cnt1);
    for(i=0;i<cnt1;i++){
        scanf("%d",&l);
        scanf("%lf",&t);
        a[l] = a[l] + t; 
    }   
    for(i=0;i<1024;i++){
        if(a[i]!= 0) 
          cnt2++;
    }
    printf("%d",cnt2);
    for(i=1023;i>=0;i--){
        if(a[i]!=0){
            printf(" %d %.1lf",i,a[i]);
        }
    }
    return 0;
}