多项式求和_系统运维

概述 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 informa

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?N?1???? N?2?? a?N?2???? ... N?K?? a?N?K????

where K is the number of nonzero terms in the polynomial, N?i?? and a?N?i???? (,) are the exponents and coefficIEnts,respectively. It is given that 1，0.

Output Specification:

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 3.22 2 1.5 1 0.5

@H_219_419@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("0\n");    }        return 0;    }