题解 | #称砝码#
称砝码
https://www.nowcoder.com/practice/f9a4c19050fc477e9e27eb75f3bfd49c
#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>
//#include <stack>
using namespace std;
vector<int> sum_stack(vector<int>& A, vector<int>& B) {
unordered_set<int> f;//维护一个无序set表
while (!A.empty()) {
int x = A.back();
A.pop_back();
//for (auto& i : B) { //似乎遍历一次,将B清空了??
for (int i = 0; i < B.size(); i++) {
int y = B[i];
int ans = x + y;
f.insert(ans);//对应插入insert
}
}
vector<int> sum;
for(int i:f){ //c++11新做法
sum.push_back(i);
//cout<<i;
}
//cout<<" ";
//B如果是栈,则只能使用一次
// while(!B.empty()){
// int y = B.back();
// B.pop_back();
// f.push_back(x+y);
// }
return sum;
}
int main() {
int n;
cin >> n;
vector<int> m(n);
for (int i = 0; i < n; i++) {
cin >> m[i];
}
vector<int> x(n);
for (int i = 0; i < n; i++) {
cin >> x[i];
}
vector<int> sum(1,0);
for (int i = 0; i < n; i++) {
//vector<int> A;
vector<int> A;
for (int j = 0; j <= x[i]; j++) {
A.push_back(m[i]*j);
//cout << (m[i]*j);
}
// cout << " ";
sum = sum_stack(sum, A);
}
cout << sum.size();
}
/* 单调栈超时
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> sum_stack(vector<int>& A, vector<int>& B) {
vector<int> f;//维护一个单调栈
while (!A.empty()) {
int x = A.back();
A.pop_back();
for (int i = 0; i < B.size(); i++) {
stack<int> stk;
int y = B[i];
int ans = x + y;
while (!f.empty() && f.back() >= ans) {
int temp = f.back();
//去重
if (temp != ans) stk.push(temp);
f.pop_back();
}
f.push_back(ans);
while (!stk.empty()) {
int temp = stk.top();
f.push_back(temp);
stk.pop();
}
}
}
//B如果是栈,则只能使用一次
// while(!B.empty()){
// int y = B.back();
// B.pop_back();
// f.push_back(x+y);
// }
return f;
}
int main() {
int n;
cin >> n;
vector<int> m(n);
for (int i = 0; i < n; i++) {
cin >> m[i];
}
vector<int> x(n);
for (int i = 0; i < n; i++) {
cin >> x[i];
}
vector<int> sum(1, 0);
for (int i = 0; i < n; i++) {
vector<int> A;
for (int j = 0; j <= x[i]; j++) {
A.push_back(m[i]*j);
//cout << (m[i]*j);
}
// cout << " ";
sum = sum_stack(sum, A);
}
cout << sum.size();
}
*/
