多项式求和

发布时间：2020-12-14 00:36:10 所属栏目：Linux 来源：网络整理

导读：? This time,you are supposed to find? A + B?where? A?and? B?are two polynomials. Input Specification: Each input file contains one test case. Each case occupies 2 lines,and each line contains the information of a polynomial: K? N ? 1 ??? a

This time,you are supposed to find?

Input Specification:

Each input file contains one test case. Each case occupies 2 lines,and each line contains the information of a polynomial:

where?

Output Specification:

For each test case you should output the sum of?

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;
struct Node{
    double ce;
    int  ex;
}n1[100],n2[100],n3[100],n4[100];
bool cmp(Node a,Node b){
    return a.ex>b.ex;
} 
int  vis[100];
int main()
{
        int  k1,k2;
        scanf("%d",&k1);
        for(int  j =0;j<k1;j++){
            scanf("%d%lf",&n1[j].ex,&n1[j].ce);
        }
        scanf("%d",&k2);
        for(int  j =0;j<k2;j++){
            scanf("%d%lf",&n2[j].ex,&n2[j].ce);
        }
        int cnt = 0;
        int  flag  ;
        for(int j=0;j<k1;j++){
            int x=n1[j].ex;
            flag  = 0;
            for(int  k =0;k<k2;k++){
                if(x==n2[k].ex){
                    n3[cnt].ex=x;
                    n3[cnt++].ce=n1[j].ce+n2[k].ce;
                    vis[x]=1;
                    flag  =1;
                    break;
                }
            }
            if(!flag){
                    n3[cnt].ex=x;
                    n3[cnt++].ce=n1[j].ce;
            }    
        }
        for(int i =0;i<k2;i++){
            int x = n2[i].ex;
            if(!vis[x]){
                n3[cnt].ex=x;
                n3[cnt++].ce=n2[i].ce;
            }
        }
        sort(n3,n3+cnt,cmp);
        int cnt1=0;
        for(int i=0;i<cnt;i++){
            if(n3[i].ce!=0){
                n4[cnt1].ex=n3[i].ex;
                n4[cnt1++].ce=n3[i].ce;
        }
        }
        if(cnt1!=0){
        printf("%d ",cnt1);
        for(int i =0;i<cnt1;i++){
            printf("%d %.1f%c",n4[i].ex,n4[i].ce,i==cnt1-1?‘n‘:‘ ‘);
        }
    }
    else{
        printf("0n");
    }
        return 0;
    
}

（编辑：李大同）

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