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
第一次写,部分案例通过:

#include<stdio.h>

int main(){
    int k,n;
    double a;
    int i;
    while(scanf("%d",&k)!=EOF){

        double arr[1001];
        int num = 0;
        for(i=0;i<1001;i++){
            arr[i] = 0;
        }
        for(i=0;i<k;i++){
            scanf("%d %lf",&n,&a);
            arr[n] = arr[n]+a;
            num++;
        }
        scanf("%d",&k);
        for(i=0;i<k;i++){
            scanf("%d %lf",&n,&a);
            if(arr[n]==0)
                num++;
            arr[n] = arr[n]+a;
        }

        printf("%d",num);
        for(i=1000;i>=0;i--){
            if(arr[i]!=0)
                printf(" %d %.1lf",i,arr[i]);
        }
        printf("\n");
    }
    return 0;
}

修改如下,num值单独计算,然后空间开辟稍大,通过了:

#include<stdio.h>

int main(){
    int k,n;
    double a;
    int i;
    while(scanf("%d",&k)!=EOF){

        double arr[1004];
        int num = 0;
        for(i=0;i<1004;i++){
            arr[i] = 0;
        }
        for(i=0;i<k;i++){
            scanf("%d %lf",&n,&a);
            arr[n] = arr[n]+a;
        }
        scanf("%d",&k);
        for(i=0;i<k;i++){
            scanf("%d %lf",&n,&a);
            arr[n] = arr[n]+a;
        }
		for(i=1000;i>=0;i--){
            if(arr[i]!=0)
                num++;
        }

        printf("%d",num);
        for(i=1000;i>=0;i--){
            if(arr[i]!=0)
                printf(" %d %.1lf",i,arr[i]);
        }
        printf("\n");
    }
    return 0;
}






1002 A+B for Polynomials 是一道编程题目,通常是在考察Java中处理多项式加法的问题。在这个问题中,你需要编写一个程序,让用户输入两个多项式的系数(如a_n*x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0的形式),然后计算它们的和,并按照同样的形式表示出来。 在Java中,你可以创建一个`Polynomial`类,包含一个数组来存储系数和最高次数的信息。用户输入的每个多项式可以被解析成这样的结构,然后通过遍历并累加系数来完成加法操作。最后,将结果转换回字符串形式展示给用户。 以下是简化版的代码示例: ```java class Polynomial { int[] coefficients; int degree; // 构造函数,初始化数组 public Polynomial(int[] coeffs) { coefficients = coeffs; degree = coefficients.length - 1; } // 加法方法 Polynomial add(Polynomial other) { Polynomial result = new Polynomial(new int[coefficients.length + other.coefficients.length]); for (int i = 0; i < coefficients.length; ++i) { result.coefficients[i] += coefficients[i]; } for (int i = 0; i < other.coefficients.length; ++i) { result.coefficients[i + coefficients.length] += other.coefficients[i]; } result.degree = Math.max(degree, other.degree); return result; } @Override public String toString() { StringBuilder sb = new StringBuilder(); if (degree >= 0) { for (int i = degree; i >= 0; --i) { sb.append(coefficients[i]).append('*x^').append(i).append(" + "); } // 移除最后一个 " + " sb.setLength(sb.length() - 2); } else { sb.append("0"); } return sb.toString(); } } // 主函数示例 public static void main(String[] args) { Polynomial poly1 = new Polynomial(...); // 用户输入第一个多项式的系数 Polynomial poly2 = new Polynomial(...); // 用户输入第二个多项式的系数 Polynomial sum = poly1.add(poly2); System.out.println("Result: " + sum); } ```
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