1002. A+B for Polynomials (25)

最新推荐文章于 2024-01-28 22:42:58 发布

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最新推荐文章于 2024-01-28 22:42:58 发布

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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A和B，输出它们相加的结果。该算法通过扫描输入的每一项，并将对应的系数累加到结果中，最终输出合并后的多项式。

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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 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

#include<bits/stdc++.h>
using namespace std;

int i,j,k,n,m;
double num[10000];

int main()
{
    int k1,k2;
    int a;double b;
    scanf("%d",&k1);
    for(int i=0;i<k1;i++){
        scanf("%d%lf",&a,&b);
        num[a]+=b;
    }
    scanf("%d",&k2);
    for(int i=0;i<k2;i++){
        scanf("%d%lf",&a,&b);
        num[a]+=b;
    }
    int sum=0;
    for(int i=0;i<=1000;i++)
        if(num[i]!=0)sum++;
    printf("%d",sum);
    if(sum>0){
        for(int i=1000;i>=0;i--)
            if(num[i]!=0)printf(" %d %.1lf",i,num[i]);
    }
    return 0;
}

#include <iostream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;
int i,j,k,n,m;
int a,b;
int flag;
int k1,k2;
struct node
{
    int no;
    double num;
};

int main()
{
    node p1[2000],p2[2000];
    node p[2000];
    p1[20].no=-1,p2[20].no=-1;
    scanf("%d",&k1);
    for(int i=0;i<k1;i++){
        scanf("%d %lf",&p1[i].no,&p1[i].num);
    }
    scanf("%d",&k2);
    for(int i=0;i<k2;i++){
        scanf("%d %lf",&p2[i].no,&p2[i].num);
    }
    i=0,j=0,k=0;
    while(1){
        if(i==k1&&j==k2)break;
        if(i==k1||j==k2){
            if(i==k1){
              p[k].no=p2[j].no;
              p[k].num=p2[j].num;
              k++,j++;
            }
            else{
               p[k].no=p1[i].no;
               p[k].num=p1[i].num;
               k++,i++;
            }
        }
        else if(p1[i].no==p2[j].no){
            p[k].no=p1[i].no;
            p[k].num=p1[i].num+p2[j].num;
            k++,j++,i++;
        }
        else if(p1[i].no>p2[j].no){
            p[k].no=p1[i].no;
            p[k].num=p1[i].num;
            k++,i++;
        }
        else if(p1[i].no<p2[j].no){
            p[k].no=p2[j].no;
            p[k].num=p2[j].num;
            k++,j++;
        }
    }
    int kk=k;
    for(int i=0;i<k;i++){
        if(p[i].num==0)kk--;
    }
    printf("%d",kk);
    for(int i=0;i<k;i++){
         if(p[i].num==0)continue;
         printf(" %d %.1f",p[i].no,p[i].num);
    }
    return 0;
}