1002. A+B for Polynomials (25)

This time, you are supposed to find A+B where A and B are two polynomials.InputEach 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, whereK 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.OutputFor 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

指数相同时系数相加，结果不为1时放入结果，指数不相同时依次放入结果中

#include <iostream>
#include <cstdio>
using namespace std;
/* A和B是两个多项式， exponents为指数,coefficients为系数
*
*
*/
struct Poly
{
int ex;
float co;
};
Poly A[10];
Poly B[10];
Poly C[20];
int main(void)
{
int m, n;
scanf("%d", &m);
for (int i = 0; i < m; ++i)
scanf("%d%f", &A[i].ex, &A[i].co);
scanf("%d", &n);
for (int i = 0; i < n; ++i)
scanf("%d%f", &B[i].ex, &B[i].co);
int i = 0, j = 0, k = 0;
while (i < m && j < n)
{
if (A[i].ex > B[j].ex)
{
C[k].ex = A[i].ex;
C[k].co = A[i].co;
++i;
++k;
}
else if(A[i].ex == B[j].ex)
{
if (A[i].co + B[j].co != 0) //注意这里一定要先判断A[i].co + B[j]是否等于0，如果等于0不用放入结果中，如果不等于0，则放入结果中
{
C[k].ex = A[i].ex;
C[k].co = A[i].co + B[j].co;
++k;
}
++i;
++j;
}
else
{
C[k].ex = B[j].ex;
C[k].co = B[j].co;
++j;
++k;
}
}
while (i < m)
{
C[k].ex = A[i].ex;
C[k].co = A[i].co;
++i;
++k;
}
while (j < n)
{
C[k].ex = B[j].ex;
C[k].co = B[j].co;
++j;
++k;
}
printf("%d", k);
for (i = 0; i < k; ++i)
printf(" %d %.1f", C[i].ex, C[i].co);//把空格的输出放到这里
printf("\n");
return 0;

}

[/code]