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甲级1002 A+B for Polynomials (25)
题目描述:
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
@:
polynomial 多项式
exponents 指数
coefficients 系数
题目要求:
给出俩个多项式的指数和系数,没有给出基数,只要将指数相同的多项式进行系数相加即可。输出时按照指数从高到低的顺序输出。
注意题意,一开始没有看题意直接看的实例,忽略了负数情况相加之后系数可能为零从而不写,所以常用的单词还是得认得。。。
代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <math.h> using namespace std; int main() { int i,j,a,n; int con=0; double s[1200],b; int vis[1200]; memset(s,0,sizeof(s)); memset(vis,0,sizeof(vis)); for(i=0;i<2;i++) { scanf("%d", &n); for(j=0;j<n;j++) { scanf("%d %lf",&a,&b); s[a]+=b; } } for(i=0;i<=1000;i++) if(s[i]!=0) con++; printf("%d", con); for(i=1000;i>=0;i--) if(s[i]!=0) printf(" %d %.1f",i,s[i]); printf("\n"); return 0; }
甲级1002 A+B for Polynomials (25)
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