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


#inc<span style="font-family: "Droid Sans Mono", Consolas, "Courier New", monospace; font-size: 1em; line-height: 1.5;">lu</span><span style="font-family: "Droid Sans Mono", Consolas, "Courier New", monospace; font-size: 1em; line-height: 1.5;">de<iostream></span>
#include<cstring>
using namespace std; int main()
{
int i, k1, k2, j,max,count;
double a[5][1005], b[1005];
while (cin >> k1)
{
max = 0;
memset(a, 0, sizeof(a));
for (i = 0; i < k1; i++)
{
cin >> j;
cin >> a[0][j];
if (max < j)
max = j;
}
cin >> k2;
for (i = 0; i < k2; i++)
{
cin >> j;
cin >> a[1][j];
if (max < j)
max = j;
}
memset(b, 0, sizeof(b));
for (j = 0; j <= 1; j++)
{ for (i = 0; i <=max; i++)
{
b[i] += a[j][i];
}
}
count = 0;
for (i = 0; i <= max; i++)
{
if (b[i] != 0)
{
count++;
}
}
cout << count;
for (i = max; i >=0; i--)
{
if (b[i] != 0)
{
printf(" %d %.1f", i, b[i]);
}
}
cout << endl; }
return 0;
}




PAT甲 1002. A+B for Polynomials (25) 2016-09-09 22:50 64人阅读 评论(0) 收藏的更多相关文章

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