这 真 TM 是个奇怪的问题 PAT 1009 Product of polynomials 求救~~~~

求救在先!!!

思想

该题目很简单, 数据结构用 下标做指数, 值做系数来存储多项式, 就行了…

然而!!

我在最后统计系数非零的项数的时候, 出现了一个我无法理解的错误!
用visited[]作为判断该项数的系数是否为0, 如果为0 则visited[i]值为0, 否则为 1;
打印的时候, 判断 visited是否为1, 若为1 则打印…第三个测试点, 竟然错了???

但是

我直接判断 data[i] 是否为0, 测试点顺利通过…无语

原题如下

---

1009 Product of Polynomials （25 分）

This time, you are supposed to find $A\times B$ 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:

$K$ N1N_1N​1​​ aN1a_{N_1}a​N​1​​​​ N2N_2N​2​​ aN2a_{N_2}a​N​2​​​​ ... NKN_KN​K​​ aNKa_{N_K}a​N​K​​​​

where $K$ is the number of nonzero terms in the polynomial, NiN_iN​i​​ and aNia_{N_i}a​N​i​​​​ ($i=1,2,\cdots ,K$) are the exponents and coefficients, respectively. It is given that $1\le K\le 10$, 0≤NK<⋯<N2<N1≤10000 \le N_K < \cdots < N_2 < N_1 \le 10000≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

For each test case you should output the product 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 up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

---

代码如下:

#include <stdio.h>
#include <stdlib.h>

int main(void){
  // 下标做指数, 值做系数
  double data1[3010], data2[3010];
  double data[3010];

  int k1, k2;
  int i, j;
  int e;
  double c;

  int es1[30], es2[30];  //存储 两行数的指数值
  int es[30], visited[3010];
  for(i = 0; i < 30; i++){
    es1[i] = -1;
    es2[i] = -1;
    es[i] = 0;
  }

  scanf("%d", &k1);
  for(i = 0; i < k1; i++){
    scanf("%d %lf", &e, &c);
    data1[e] = c;
    es1[i] = e;
  }

  scanf("%d", &k2);
  for(i = 0; i < k2; i++){
    scanf("%d %lf", &e, &c);
    data2[e] = c;
    es2[i] = e;
  }

  for(i = 0; i < 3010; i++){
    data[i] = 0;
    visited[i] = 0;

  }

  int t = 0;
  int count = 0;
  for(i = 0; i < k1; i++){
    for(j = 0; j < k2; j++){
        es[t] += es1[i] + es2[j];
        data[es[t]] += data1[es1[i]] * data2[es2[j]];
        //if(visited[es[t]]&&data[es[t]]!=0){
        //    count ++;
        visited[es[t]] = 1;
        //}
        if(data[es[t]] == (double)0){
          //count--;
          visited[es[t]] = 0;
        }
        t++;
    }
  }
  count=0;
  for(i=0;i<3010;i++){
    if(data[i] !=0)
      count++;
  }
  printf("%d", count);
  for(i = 3009; i >= 0; i--){
    if(data[i] != (double)0 ){   // j就是这里 !!!!!!!换成 visited, 第三个测试点过不去
        printf(" %d %.1lf", i, data[i] );
        
    }
  }



  return 0;
}