1002. A+B for Polynomials（25）

This time，you are supposed to find A+B where A+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 nonezero 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 asthe 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：多项式合并同类项，按输入格式输出最后多项式的项数、各项的指数和系数。

分析：构造一个整型数组存放多项式的系数，输入时将相同指数的系数累加，如果累加之前数组存放的值为0，那么多项式的项数加1；如果累加之后等于0，那么多项式的项数减一。注意：多项式的指数都是整数，不要瞎想了~

#include <iostream>

using namespace std;

double s[1005]={0};

int main() {

    int k,n,time=2,count=0;

    double ak;

    while(time--){

        scanf("%d",&k);

        while(k--){

            scanf(" %d %lf",&n,&ak);

            if(s[n]==0)count++;

            s[n]+=ak;

            if(s[n]==0)count--;

        }

    }

    printf("%d",count);

    for(int i=1005;i>=0;i--){

        if(s[i]!=0){

            printf(" %d %.1lf",i,s[i]);

        }

    }

    printf("\n");

    return 0;

}

巴特西

1002. A+B for Polynomials（25）—PAT 甲级

Input##

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