模拟一下就行,运用两个字符串来反复模拟。
#include <iostream> #include <queue> #include <map> #include <set> #include <cmath> #include <cstring> #include <algorithm> #include <iomanip> #include <stack> #include <numeric> #include <ctime> #include <string> #include <bitset> #include <unordered_map> #include <unordered_set> using namespace std; using ll = long long; const ll N = 1e6 + 5, mod = 1e9 + 7, inf = 0x3f3f3f3f; int n, q; void solve() { cin >> n >> q; string a; cin >> a; a = "*" + a; string b; while (q--) { int l, r; cin >> l >> r; b = ""; for (int i = 0; i < a.size(); i++) { b += a[i]; if (i >= l && i <= r)b += a[i]; } a = b; } a.erase(0, 1); //删除一开始加入的* cout << a << '\n'; } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t = 1; //cin>>t; while (t--) { solve(); } return 0; }