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
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1 2 |
2 1 2.4 0 3.2 2 2 1.5 1 0.5 |
Sample Output
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1 |
3 2 1.5 1 2.9 0 3.2 |
题目同样没有什么难度，对应项系数相加即可。然而，两个多项式的非零项个数均大于零，但是最后结果多项式的非零项可以为零个，这一点可能被忽视，其他的也没有什么坑了。
代码如下：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 |
#include<stdio.h> double coefficient[1001]; int main() { int i,count=0,n,index; double temp; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d%lf",&index,&temp); coefficient[index]+=temp; } scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d%lf",&index,&temp); coefficient[index]+=temp; } for(i=1000;i>=0;i--) { if(coefficient[i]!=0) { count++; index=i; } } if(count>0) { printf("%d ",count); for(i=1000;i>index;i--) { if(coefficient[i]!=0) printf("%d %.1lf ",i,coefficient[i]); } printf("%d %.1lf",i,coefficient[i]); } else printf("%d",count); } |