甲级1002 A+B for Polynomials (25)

题目描述：

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

@：
polynomial    多项式

exponents     指数

coefficients  系数

题目要求：

给出俩个多项式的指数和系数，没有给出基数，只要将指数相同的多项式进行系数相加即可。输出时按照指数从高到低的顺序输出。

注意题意，一开始没有看题意直接看的实例，忽略了负数情况相加之后系数可能为零从而不写，所以常用的单词还是得认得。。。

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    int i,j,a,n;
    int con=0;
    double s[1200],b;
    int vis[1200];
    memset(s,0,sizeof(s));
    memset(vis,0,sizeof(vis));

    for(i=0;i<2;i++)
    {
        scanf("%d", &n);
        for(j=0;j<n;j++)
        {
            scanf("%d %lf",&a,&b);
            s[a]+=b;
        }
    }

    for(i=0;i<=1000;i++)
        if(s[i]!=0)
        con++;

    printf("%d", con);

    for(i=1000;i>=0;i--)
        if(s[i]!=0)
            printf(" %d %.1f",i,s[i]);

    printf("
");

    return 0;
}