士兵排列问题

解法一：

deque实现队头入队和队尾入队即可得到编号排列，每个士兵有二个属性：编号、能力值。

#include<iostream>
#include<algorithm>
#include<deque>
#include<vector>
using namespace std;
#define endl '\n'
struct shib {int bianh;int nengl;
};
bool cmp(struct shib& a, struct shib& b) {return a.nengl > b.nengl;}
void solve() {deque<int> p;int n, a;cin >> n;string st;cin >> st;for (int i = 0; i < st.size(); i++) {if (st[i] == '0') p.push_front(i + 1);else p.push_back(i + 1);}vector<struct shib> vec(n);for (int i = 0; i < n; i++) {vec[i].bianh = p[i];cin >> a;vec[i].nengl = a;}sort(vec.begin(), vec.end(), cmp);for (int i = 0; i < n; i++) {cout << vec[i].bianh << " ";}cout << endl;
}
int main() {solve();return 0;
}

解法二：感谢dalao的分享

再来一篇迭代器遍历的

#include <iostream>
#include<algorithm>
#include<string>
#include<deque>
using namespace std;
struct arm {int located;int ability;
}ar[100005];
bool compare(const arm& a, const arm& b) {return a.ability > b.ability;
}
int main()
{int t, a;deque<int> d;string str;cin >> t >> str;for (int i = 0; i < t; i++) {if (str[i] == '1') d.push_back(i+1);else d.push_front(i+1);}int i = 0;for (auto it = d.begin(); it != d.end(); it++) {ar[i++].located = *it;}for (int i = 0; i < t; i++) {cin >> a;ar[i].ability = a;}sort(ar, ar + t, compare);for (int i = 0; i < t; i++)cout << ar[i].located << " ";cout << endl;
}

关于下标在stl中的应用局限。