1002. A+B for Polynomials
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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 Input2 1 2.4 0 3.22 2 1.5 1 0.5
Sample Output3 2 1.5 1 2.9 0 3.2
解题思路:题目很简单,大意就是多项式相加,给你两个多项式的指数和系数,要你算出相加的结果
代码如下:
#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int main(){ int n,m,x,b[1001]; double y,a[1001]={0};///a数组下标相当于指数值 scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d %lf",&x,&y); a[x]=y; } scanf("%d",&m); for(int i=0;i<m;i++) { scanf("%d %lf",&x,&y); a[x]+=y; } int t=0; for(int i=0;i<1001;i++) if(a[i]) b[t++]=i; printf("%d",t); for(int i=t-1;i>=0;i--) printf(" %d %.1lf",b[i],a[b[i]]); printf("\n"); //printf("%d %.1lf\n",b[0],a[b[0]]); return 0;}
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