PAT甲级真题打卡:1002. A+B for Polynomials

题目:

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

参考代码:

#include <stdio.h>
#include <string.h>
using namespace std;

double a[1002];
double b[1002];


int main(void) {
int n, i, temp;
memset(a,0, sizeof(a));
memset(b,0, sizeof(b));
scanf("%d", &n);
for(i = 0; i <= n; i++) {
scanf("%d", &temp);
scanf("%lf", &a[temp]);
}
scanf("%d", &n);
for(i = 0; i < n; i++) {
scanf("%d", &temp);
scanf("%lf", &b[temp]);
}
int count = 0;
for (i = 0; i < 1002; i++) {
a[i] += b[i];
if(a[i] != 0)
count++;
}
printf("%d",count);
for(i = 1001; i >= 0; i--) {
if(a[i] != 0) {
count--;
printf(count==0?" %d %.1f\n":" %d %.1f", i, a[i]);
}
}
return 0;
}

分析:

这是一个比较简单的多项式题目,只要看懂了题就可以迎刃而解。这题关键在于输入输出。

1.不能有空格

2.多项式相加系数为0时如何处理。

 

posted on 2018-05-02 19:36  唐宇devpro  阅读(165)  评论(0编辑  收藏  举报

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