PAT (Advanced Level) Practise - 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


多项式求和。

#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <string.h>
using namespace std;
#define LL long long
const int maxn = 1111;
const int inf = 0x3f3f3f3f;
double a[maxn];

void doit(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
int k;
double s;
cin>>k>>s;
a[k]+=s;
}
}
int main()
{
memset(a,0,sizeof a);
doit();
doit();
int num=0;
for(int i=1000;i>=0;i--){
if(a[i]!=0){
num++;
}
}
printf("%d",num);
for(int i=1000;i>=0;i--){
if(a[i]!=0){
printf(" %d %.1lf",i,a[i]);
}
}
printf("\n");
return 0;
}