pat甲级1002

1002. A+B for Polynomials (25)

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

1 #include<cstdio>
2 struct node
3 {
4     int e;//指数
5     double cof;//系数
6 }a[1500],b[1500];
7 double ans[2002]={0};
8 int main()
9 {
10     int m,n,sum=0;
11     scanf("%d",&m);
12     for(int i=0;i<m;i++)
13     {
14         scanf("%d %lf",&a[i].e,&a[i].cof);
15         ans[a[i].e]+=a[i].cof;
16     }
17     scanf("%d",&n);
18     sum=m+n;
19     for(int i=0;i<n;i++)
20     {
21         scanf("%d %lf",&b[i].e,&b[i].cof);
22         ans[b[i].e]+=b[i].cof;
23     }
24     for(int i=0;i<m;i++)
25     {
26         for(int j=0;j<n;j++)
27         {
28             if(a[i].e==b[j].e)
29                 sum--;
30         }
31     }
32
33     printf("%d",sum);
34     for(int i=2000;i>=0;i--)
35     {
36         if(ans[i]!=0)
37             printf(" %d %.1lf",i,ans[i]);
38     }
39
40
41     return 0;
42 }

有四个测试点过不去，应该是最后计含有非零项多项式时候少考虑了一种情况——两个指数相同的项的系数和为0

#include<cstdio>
struct node
{
int e;//指数
double cof;//系数
}a[1500],b[1500];
double ans[2002]={0};
int main()
{
int m,n,sum=0;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d %lf",&a[i].e,&a[i].cof);
ans[a[i].e]+=a[i].cof;
}
scanf("%d",&n);
sum=m+n;
for(int i=0;i<n;i++)
{
scanf("%d %lf",&b[i].e,&b[i].cof);
ans[b[i].e]+=b[i].cof;
}
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(a[i].e==b[j].e)
sum--;
if(a[i].e==b[j].e&&a[i].cof+b[j].cof==0)
sum--;
}
}

printf("%d",sum);
for(int i=2000;i>=0;i--)
{
if(ans[i]!=0)
printf(" %d %.1lf",i,ans[i]);
}

return 0;
}

　改动一段代码果然解决了问题