1002.A+B for Polynomials (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1002

题目：

  
  
  
  

   
   
   
   
 
    

      时间限制 
    
 
    

      400 ms 
     
 
    
 
   
   
   
   
   
 
    

      内存限制 
    
 
    

      65536 kB 
     
 
    
 
   
   
   
   
   
 
    

      代码长度限制 
    
 
    

      16000 B 
     
 
    
 
   
   
   
   
   
 
    

      判题程序 
    
 
    
 
     Standard 
    
 
   
   
   
   
   

     作者 
   
   
   
   
   

     CHEN, Yue 
    
 
   
  
  
  
  
  
  
  
  

   
   
   
   
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

   
   
   
   
 
    
 
   
  
  
  
  

分析：

这是一道多项式运算的题目，看起来不难，但是有一些注意点：

1）要有分别保存多项式系数和指数的数组

2）对于运算，注意有运算后为0的情况，所以总的个数会减少

3）同时，也有两个多项式全部抵消的情况，即0项结果，那么注意这时要输出的只能是0，而不能是0空格。

以下是我没有注意到格式方面的错误提交后的截图

AC的代码如下：

#include<iostream>
#include<string.h>
using namespace std;
int a[30];//存放指数
double b[30];//存放系数
int main(void){
 int i,j,n,temp,flag;
 double temp2;
 bool added;//如果指数相同则系数相加减
 memset(a,0,sizeof(a));
 memset(b,0,sizeof(b));//init()
 scanf("%d",&n);
 for(i = 0; i < n; i++){
  scanf("%d",&a[i]);
  scanf("%lf",&b[i]);
 }
 flag = n;
 scanf("%d",&n);
 for(i = 0; i < n; i++){
  added = false;
  scanf("%d",&temp);
  scanf("%lf",&temp2);
  for(j = 0; j < flag; j++){
   if(temp == a[j]){//如果指数相同则系数相运算
    b[j] += temp2;
    added = true;
   }
  }
  if(added == false){//否则就把新的项存下来
   a[flag] = temp;
   b[flag] = temp2;
   flag ++;
  }
 }
 int count = 0;
 for(i = 0; i < flag; i++){//统计指数相同后系数变为0即抵消的项数目
  if(b[i] == 0)
   count++;
 }
 printf("%d", flag - count);//得到最后的有效的项数目
 for(i = 0; i < flag-1; i++){//按指数大小排序
  for(j = i + 1; j < flag; j++){
   if(a[j] > a[i]){
    temp = a[i];
    a[i] = a[j];
    a[j] = temp;
    temp2 = b[i];
    b[i] = b[j];
    b[j] =temp2;
   }
  }
 }
 while(b[flag-1] == 0){
  flag --;
 }
 for(i = 0; i < flag; i++){//格式化输出
  if(b[i] != 0){
   if(i != flag - 1){
    printf(" %d %.1lf",a[i],b[i]);
   }
   else {
    printf(" %d %.1lf\n",a[i],b[i]);
   }
  }
 }
// system("pause");
 return 0;
}

AC截图如下：

当然，这是很早以前写的啦，会有繁琐的地方。

也顺便把王道论坛的答案1002部分放上来吧

#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;
}

——Apie陈小旭