1002 A+B for Polynomials (25分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

this time, you are supposed to find A+B where A and B are two polynomials.

Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
KN1aN1N2aN2…NKaNKK N1 aN1 N2 aN2 … NK aNKKN1aN1N2aN2…NKaNK where K is the number of nonzero terms in the polynomial, NiNiNi and aNi(i=1,2,⋯,K)aNi (i=1,2,⋯,K)aNi(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101≤K≤101≤K≤10，0≤NK<⋯<N2<N1≤10000≤NK<⋯<N2<N1≤10000≤NK<⋯<N2<N1≤1000.

Output Specification:
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

注意要点：
1、如果相加后总多项式为0，则输出0
2、如果某一项为0，则不输出

#include<bits/stdc++.h>
using namespace std;
double s[1010];
int main() {int k1; cin >> k1; while(k1--) {int x; double d; cin >> x >> d;s[x] += d;}cin >> k1; while(k1--) {int x; double d; cin >> x >> d;s[x] += d;}int num = 0;for(int i = 1009; i >= 0; i--) if(s[i] != 0) num++; if(num == 0) cout << 0;else {cout << num;for(int i = 1009; i >= 0; i--) {if(s[i] != 0) printf(" %d %.1lf", i, s[i]);}}putchar('\n');return 0;
}

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1002 A+B for Polynomials (25分)_29行代码AC

立志用最少的代码做最高效的表达

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