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题意：
四个数字，可以改变顺序，中间可以添加+,-,*符号。数字之间还可以合并成一个数。
求最终能生成多少个数字。

思路：
暴力枚举生成了哪几个数，再枚举这几个数之间符号是啥，再用栈算出这个结果，用set去重。
写的贼麻烦，快麻了。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5+10;
set<int>st;
int get(deque<pair<int,int>>que) {deque<pair<int,int>>d = que;deque<pair<int,int>>now;while(!que.empty()) {pair<int,int>num = que.front();que.pop_front();if(num.first == 1 && num.second == 2) {pair<int,int>nex = que.front();que.pop_front();pair<int,int>pre = now.back();now.pop_back();int number = nex.second * pre.second;now.push_back({0,number});} else {now.push_back(num);}}que = now;now.clear();while(!que.empty()) {pair<int,int>num = que.front();que.pop_front();if(num.first == 1) {if(num.second == 0) {pair<int, int> nex = que.front();que.pop_front();pair<int, int> pre = now.back();now.pop_back();int number = nex.second + pre.second;now.push_back({0,number});} else if(num.second == 1) {pair<int, int> nex = que.front();que.pop_front();pair<int, int> pre = now.back();now.pop_back();int number = pre.second - nex.second;now.push_back({0,number});}} else {now.push_back(num);}}return now.front().second;
}
void fuhao(int id,vector<int>v,deque<pair<int,int>>que) {if(id == v.size() - 1) {que.push_back({0,v[id]});int num = get(que);if(num >= 0) {st.insert(num);}return ;}que.push_back({0,v[id]});for(int i = 0;i < 3;i++) { //+ - *que.push_back({1,i});fuhao(id + 1,v,que);que.pop_back();}
}
void cal(vector<int>v) {if(v.size() == 1) return;deque<pair<int,int>>que;fuhao(0,v,que);
}
vector<vector<int>>vec;
int a[10];
void dfs(int id,int now,vector<int>v) {if(id == 5) {if(now) v.push_back(now);vec.push_back(v);return;}if(now) v.push_back(now);dfs(id + 1,a[id],v);if(now) v.pop_back();now = now * 10 + a[id];dfs(id + 1,now,v);
}
int main() {scanf("%d%d%d%d",&a[1],&a[2],&a[3],&a[4]);sort(a + 1,a + 1 + 4);do {vector<int> v;dfs(1, 0, v);}while(next_permutation(a + 1,a + 1 + 4));for(int i = 0;i < vec.size();i++) {cal(vec[i]);}printf("%d\n",st.size());return 0;
}