Contents

1. 1. Input Specification:
2. 2. Output Specification:
3. 3. Sample Input:
4. 4. Sample Output:
5. 5. 题目大意：
6. 6. 分析：

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 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 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 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

题目大意：

给出两行，分别代表两个多项式，要求合并。每一行第一个数字表示该多项式的项数n，其后n对多项式。每对前一个未知数的指数，后一个为该未知数的系数。合并结果的系数部分精确到小数点后一位。

分析：

题目给出了多项式指数的范围，0~1000，可以用数组dict[1010]记录每一个多项式的系数。遍历dict[1000] ~ dict[0]，当其中值不为0的项是符合要求的项，按照要求输出即可。

#include<iostream>
using namespace std;
int main(){
    int m,n,a,cnt=0;
    double b,dict[1010]={0};
    scanf("%d",&m);
    for (int i=0;i<m;i++){
        scanf("%d%lf",&a,&b);
        dict[a] += b;
    }
    scanf("%d",&n);
    for (int i=0;i<n;i++){
        scanf("%d%lf",&a,&b);
        dict[a] += b;
    }
    for (int i=1000;i>=0;i--)
        if (dict[i] != 0.0) cnt++;
    printf("%d",cnt);
    for (int i=1000;i>=0;i--)
        if (dict[i] != 0.0) printf(" %d %.1f",i,dict[i]);
    return 0;
}