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;
}

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