1002. A+B for Polynomials (25)

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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 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
解题思路：
1.哈希法保存两多项式。并对其做标记
2.根据标记进行加法，对结果进行哈希保存，并做标记。要注意和为零的情况。
3.从数组尾部根据标记遍历数组，保证从大到小输出。
示例代码：

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<cstdio>using namespace std;const int MAX = 1002;bool flag1[MAX];bool flag2[MAX];bool flag3[MAX];int main(){    float a[MAX];    float b[MAX];    int n1,n2,exp,n3=0;    float coe;    cin >> n1;    for (int i = 0; i < n1; i++)    {        cin >> exp >> coe;        a[exp] = coe;        flag1[exp] = true;    }    cin >> n2;    for (int i = 0; i < n2; i++)    {        cin >> exp >> coe;        b[exp]= coe;        flag2[exp] = true;    }    float sum[MAX];    for (int i = 0; i < MAX; i++)    {        if (flag1[i] && flag2[i])        {            sum[i] = a[i] + b[i];        }        else if (flag1[i])        {            sum[i] = a[i];        }        else if (flag2[i])        {            sum[i] = b[i];        }        if (!(flag1[i] == false && flag2[i] == false))        {            if (sum[i] != 0.0f)            {                flag3[i] = true;                n3++;            }        }    }    cout << n3;    for (int i = MAX - 1; i >= 0; i--)    {        if (flag3[i])            printf(" %d %.1f",i,sum[i]);    }    system("pause");    return 0;}