PAT 甲级测试题目 -- 1009 Product of Polynomials

标签：==   空格   turn   ret   namespace   code   str   多项式   product   

题目链接
这题占个坑吧。。。我用了两种思路，一种将保存结果的数组初始化为 0，把乘积加上去，顺便记录下最后一个指数的值。用两次 2000 次 for 循环过滤掉 0 值以及输出答案，这个代码过了，但是后期测试的时候，有些测试用例会多出来空格，有些则格式不对。。但是过了。。。
代码如下

#include<iostream>
#include<stdio.h>
using namespace std;
int main() {
    double A[1001], B[1001];
    int ExpA[11], ExpB[11];
    double Result[2001];
    double coe;
    int Acounts, Bcounts, exp, EndIndex = 0;

    // 初始化多项式


    for (int i = 0; i < 1001; i++) {
        A[i] = B[i] = 0;
    }
    for (int i = 0; i < 2001; i++) {
        Result[i] = 0;
    }

    cin >> Acounts;
    for (int i = 0; i < Acounts; i++) {
        cin >> exp >> coe;
        ExpA[i] = exp;
        A[exp] = coe;
    }
    cin >> Bcounts;
    for (int i = 0; i < Bcounts; i++) {
        cin >> exp >> coe;
        ExpB[i] = exp;
        B[exp] = coe;
    }

    for (int i = 0; i < Acounts; i++) {
        for (int j = 0; j < Bcounts; j++) {
            //cout << "=====================" << endl;
            //cout << "Result[ExpA[i] + ExpB[j]] = " << Result[ExpA[i] + ExpB[j]] << endl;
            Result[ExpA[i] + ExpB[j]] += A[ExpA[i]] * B[ExpB[j]];
            //cout << "A[ExpA[i]] * B[ExpB[j]] = " << A[ExpA[i]] * B[ExpB[j]] << endl;
            //cout << "Result[ExpA[i] + ExpB[j]] = " << Result[ExpA[i] + ExpB[j]] << endl;
            //cout << "=====================" << endl;
            EndIndex = ExpA[i] + ExpB[j];
        }
    }

    int AllResCount = 0;

    for (int i = 2000; i >= 0; i--) {
        if (Result[i] != 0) {
            AllResCount++;
        }

    }

    cout << AllResCount << " ";
    for (int i = 2000; i >= 0; i--) {
        if (Result[i] != 0) {
            printf("%d %.1f", i, Result[i]);
            if (i != EndIndex)
                cout << " ";

        }

    }

    return 0;
}

另外一种思路是将保存结果的数组初始化为 -1 ，边计算多项式边记录非零多项式的项数，最后输出答案，测试点 0 答案错误。。。我百度了好久，好多帖子指出是跟系数为 0 有关，我自己做了几个系数为 0 测试用例，测试结果和我在纸上计算的答案一致，包括格式也正确，我实在弄不懂了。。。代码如下：

#include<iostream>
#include<stdio.h>
using namespace std;
int main() {
    double A[1001], B[1001];
    int ExpA[11], ExpB[11], ItemIndex[10];
    double Result[2001];
    double coe;
    int Acounts, Bcounts, exp, ResCounts = 0;

    // 初始化多项式


    for (int i = 0; i < 1001; i++) {
        A[i] = B[i] = -1;
    }
    for (int i = 0; i < 2001; i++) {
        Result[i] = -1;
    }

    cin >> Acounts;
    for (int i = 0; i < Acounts; i++) {
        cin >> exp >> coe;
        ExpA[i] = exp;
        A[exp] = coe;
    }
    cin >> Bcounts;
    for (int i = 0; i < Bcounts; i++) {
        cin >> exp >> coe;
        ExpB[i] = exp;
        B[exp] = coe;
    }

    int indexCount = 0;
    for (int i = 0; i < Acounts; i++) {     
        for (int j = 0; j < Bcounts; j++) {         
            if (Result[ExpA[i] + ExpB[j]] == -1) {
                Result[ExpA[i] + ExpB[j]] = A[ExpA[i]] * B[ExpB[j]];

                if (Result[ExpA[i] + ExpB[j]] != 0) {
                    ResCounts++;
                }
            }
            else {
                if (Result[ExpA[i] + ExpB[j]] == 0 && A[ExpA[i]] * B[ExpB[j]] != 0)
                    ResCounts++;
                Result[ExpA[i] + ExpB[j]] += A[ExpA[i]] * B[ExpB[j]];
            }
            
        }
    }

    cout << ResCounts << " ";
    for (int i = 2000; i >= 0; i--) {
        
        if (Result[i] != -1 && Result[i] != 0) {
            ResCounts--;
            //cout << "----" << ResCounts << "----" << endl;
            printf("%d %.1f", i, Result[i]);
            if (ResCounts != 0)
                printf(" ");
        }
    }

    return 0;
}

希望有大佬可以指出错误QAQ

PAT 甲级测试题目 -- 1009 Product of Polynomials

标签：==   空格   turn   ret   namespace   code   str   多项式   product   

原文地址：https://www.cnblogs.com/Breathmint/p/10283356.html