1009.Product of Polynomials (25)

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pat-al-1009

2017-02-15

哈希表的应用
坑见注释

/** * pat-al-1009 * 2017-02-15 * C version * Author: fengLian_s */#include<stdio.h>int main(){  double hs1[1001] = {0}, hs2[1001], result[2001] = {0};//坑：要记得给hs1初始化，不然会不知道数组里存的到底是什么鬼  freopen("in.txt", "r", stdin);  int k1, k2;  scanf("%d", &k1);  for(int i = 0;i < k1;i++)  {    int tmpE;    double tmpC;    scanf("%d%lf", &tmpE, &tmpC);    //printf("tmpE = %d, tmpC = %.1lf\n", tmpE, tmpC);    hs1[tmpE] = tmpC;  }  scanf("%d", &k2);  for(int i = 0;i < k2;i++)  {    int tmpE;    double tmpC;    scanf("%d%lf", &tmpE, &tmpC);    //printf("tmpE = %d, tmpC = %.1lf\n", tmpE, tmpC);    for(int j = 0;j <= 1000;j++)    {      if(hs1[j]*tmpC != 0)      {        //printf("hs1[%d] = %.1lf, tmpC = %.1lf\n", j, hs1[j], tmpC);        result[tmpE+j] += hs1[j]*tmpC;      }    }    hs2[tmpE] = tmpC;  }  int cnt = 0;  for(int i = 0;i <= 2000;i++)  {    if(result[i] != 0)      cnt++;  }  printf("%d", cnt);  for(int i = 2000;i >= 0;i--)  {    if(result[i] != 0)      printf(" %d %.1lf", i, result[i]);  }  putchar('\n');}

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