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
#include
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
#include
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
#include
我将之前的代码用map改写之后的代码
#include
#include
#include
#include
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
#include
#include
#include
using namespace std;
struct Decline {
bool operator()(const int x1, const int x2) const {
return x1 > x2;
}
};
int main() {
map A, B, Result;
int Exp, number;
double Coe;
cin >> number;
while (number--) {
cin >> Exp >> Coe;
A.insert(pair(Exp, Coe));
}
cin >> number;
while (number--) {
cin >> Exp >> Coe;
B.insert(pair(Exp, Coe));
}
pair::iterator, bool> InsertFlag;
for (map::iterator i = A.begin(); i != A.end(); i++) {
for (map::iterator j = B.begin(); j != B.end(); j++) {
if (i->second * j->second != 0) {
InsertFlag = Result.insert(pair(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::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::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::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
#include
#include
#include
using namespace std;
int main() {
map A, B, Result;
int Exp, number;
double Coe;
cin >> number;
while (number--) {
cin >> Exp >> Coe;
A.insert(pair(Exp, Coe));
}
cin >> number;
while (number--) {
cin >> Exp >> Coe;
B.insert(pair(Exp, Coe));
}
pair::iterator, bool> InsertFlag;
for (map::iterator i = A.begin(); i != A.end(); i++) {
for (map::iterator j = B.begin(); j != B.end(); j++) {
if (i->second * j->second != 0) {
InsertFlag = Result.insert(pair(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::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。