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

由 不想你离开。 提交于 2020-03-07 01:54:55

题目链接

坑点：注意系数可能为负数！！只有测试点 0 过不去可能就是因为这个！！！

我最初在做这道题的时候用了两种思路，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;
}

后来通过跟大佬交流之后，发现了一些端倪
大佬的代码：

#include <iostream>
#include <map>

using namespace std;

int main()
{
    using map_t = map<int, float>;
    map_t data[2];
    int c;
    int k;
    float v;
    for (int i = 0; i < 2; ++i) {
        cin >> c;
        for (int j = 0; j < c; ++j) {
            cin >> k >> v;
            data[i][k] = v;
        }
    }

    map_t out;
    for (map_t::iterator it0 = data[0].begin(); it0 != data[0].end(); ++it0) {
        for (map_t::iterator it1 = data[1].begin(); it1 != data[1].end(); ++it1) {
            int key = it0->first + it1->first;
            out[key] += it0->second * it1->second;
            if (out[key] == 0.0f) {
                out.erase(key);
            }
        }
    }

    int size = out.size();
    printf("%d ", size);
    for (map_t::reverse_iterator it = out.rbegin(); it != out.rend(); ++it) {
        printf("%d %.1f", it->first, it->second);
        if (--size > 0) {
            printf(" ");
        }
    }

    return 0;
}

我将之前的代码用map改写之后的代码

#include<iostream>
#include<string>
#include<stdio.h>
#include<map>
using namespace std;

struct Decline {
    bool operator()(const int x1, const int x2) const {
        return x1 > x2;
    }

};

int main() {
    map<int, double, Decline> A, B, Result;
    int Exp, number;
    double  Coe;

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        A.insert(pair<int, double>(Exp, Coe));
    }

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        B.insert(pair<int, double>(Exp, Coe));
    }

    pair<map<int, double>::iterator, bool> InsertFlag;

    for (map<int, double>::iterator i = A.begin(); i != A.end(); i++) {
        for (map<int, double>::iterator j = B.begin(); j != B.end(); j++) {
            if (i->second * j->second != 0) {
                InsertFlag = Result.insert(pair<int, double>(i->first + j->first, i->second * j->second));
                if (!InsertFlag.second) {
                    Result[i->first + j->first] += i->second * j->second;
                }
            }
                
        }
    }

    int tag = Result.size();
    cout << tag << " ";
    for (map<int, double>::iterator i = Result.begin(); i != Result.end(); i++) {
        tag--;
        printf("%d %.1f", i->first, i->second);     
        if (tag != 0)
            cout << " ";
    }
    return 0;
}

这里可以发现，大佬在处理数据的时候，将加和为 0 的项目用 erase() 函数擦除了。后来经过交流发现，测试点 0 没通过的情况可能是 测试点 0 有负数，
我在对数据处理的时候，没有考虑负数加和，因此测试点 0 过不去。
所以在39行代码下方加入以下判断：

                if (Result[i->first + j->first] == 0)
                    Result.erase(i->first + j->first);

即可满分通过
完整代码如下：

#include<iostream>
#include<string>
#include<stdio.h>
#include<map>
using namespace std;

struct Decline {
    bool operator()(const int x1, const int x2) const {
        return x1 > x2;
    }

};

int main() {
    map<int, double, Decline> A, B, Result;
    int Exp, number;
    double  Coe;

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        A.insert(pair<int, double>(Exp, Coe));
    }

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        B.insert(pair<int, double>(Exp, Coe));
    }

    pair<map<int, double>::iterator, bool> InsertFlag;

    for (map<int, double>::iterator i = A.begin(); i != A.end(); i++) {
        for (map<int, double>::iterator j = B.begin(); j != B.end(); j++) {
            if (i->second * j->second != 0) {
                InsertFlag = Result.insert(pair<int, double>(i->first + j->first, i->second * j->second));
                if (!InsertFlag.second) {
                    Result[i->first + j->first] += i->second * j->second;
                }
                if (Result[i->first + j->first] == 0)
                    Result.erase(i->first + j->first);
            }
                
        }
    }

    int tag = Result.size();
    cout << tag << " ";
    for (map<int, double>::iterator i = Result.begin(); i != Result.end(); i++) {
        tag--;
        printf("%d %.1f", i->first, i->second);     
        if (tag != 0)
            cout << " ";
    }
    return 0;

}

另外，如果觉得额外写一个结构体用于降序排序有些麻烦的话，可以使用 reverse_iterator，只需要将下面的代码

        for (map<int, double>::iterator i = Result.begin(); i != Result.end(); i++) {
        tag--;
        printf("%d %.1f", i->first, i->second);     
        if (tag != 0)
            cout << " ";
    }

改成这样子

    for (map<int, double>::reverse_iterator i = Result.rbegin(); i != Result.rend(); i++) {
        tag--;
        printf("%d %.1f", i->first, i->second);     
        if (tag != 0)
            cout << " ";
    }

即可，且无需定义并实现用于降序的结构体，定义类型的时候也不需要将结构体引入泛型，完整代码如下：

#include<iostream>
#include<string>
#include<stdio.h>
#include<map>
using namespace std;
int main() {
    map<int, double> A, B, Result;
    int Exp, number;
    double  Coe;

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        A.insert(pair<int, double>(Exp, Coe));
    }

    cin >> number;
    while (number--) {
        cin >> Exp >> Coe;
        B.insert(pair<int, double>(Exp, Coe));
    }

    pair<map<int, double>::iterator, bool> InsertFlag;

    for (map<int, double>::iterator i = A.begin(); i != A.end(); i++) {
        for (map<int, double>::iterator j = B.begin(); j != B.end(); j++) {
            if (i->second * j->second != 0) {
                InsertFlag = Result.insert(pair<int, double>(i->first + j->first, i->second * j->second));
                if (!InsertFlag.second) {
                    Result[i->first + j->first] += i->second * j->second;
                }
                if (Result[i->first + j->first] == 0)
                    Result.erase(i->first + j->first);
            }
                
        }
    }

    int tag = Result.size();
    cout << tag << " ";
    for (map<int, double>::reverse_iterator i = Result.rbegin(); i != Result.rend(); i++) {
        tag--;
        printf("%d %.1f", i->first, i->second);     
        if (tag != 0)
            cout << " ";
    }
    return 0;   
}

希望能帮到大家。

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