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 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

AC代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;

double poly[3][1010];
int main(int argc, const char * argv[]) {
    int k, count = 0;
    int exp;
    double coe;
    for(int i = 0; i < 2; i++) {
        scanf("%d", &k);
        for(int j = 0; j < k; j++) {
            scanf("%d%lf", &exp, &coe);
            poly[i][exp] = coe;
        }
    }
    for(int i = 0; i < 1010; i++) {
        poly[2][i] = poly[0][i] + poly[1][i];
    }
    for(int i = 0; i < 1010; i++){
        if(poly[2][i] != 0) count++;
    }
    printf("%d", count);
    if(count) {
        for(int i = 1009; i >= 0; i--) {
            if(poly[2][i] != 0) printf(" %d %.1f", i, poly[2][i]);
        }
    }
    return 0;
}

另一种解法

//
//  main.cpp
//  pat
//
//  Created by yaojies on 16/8/16.
//  Copyright © 2016年 yaojies. All rights reserved.
//  1002. A+B for Polynomials

#include <iostream>
#include <iomanip>
using namespace std;

int main(int argc, const char * argv[]) {
    // insert code here...
    float an[3][1001]={0};
    for(int i=0;i<2;i++){
        int k;
        cin >> k;
        for(;k>0;k--){
            int n;
            float ank;
            cin >> n >> ank;
            an[i][n]=ank;
        }
    }
    int newk = 0;
    for(int j = 1000;j>=0;j--){
        an[2][j]=an[0][j]+an[1][j];
        if(an[2][j]!=0) newk++;
    }
    cout << newk;
    for(int j = 1000;j>=0;j--){
        if(an[2][j]!=0){
            //控制输出小数位数
            cout << " " << j << " " << fixed << setprecision(1) << an[2][j];
        }
    }
    cout << endl;
    return 0;
}