由于每个串的长度为5000，我们去枚举两个自串的对称点（这里注意一下，枚举的时候有两种情况的区间），然后用尺取法爬一遍。
ac代码：
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<algorithm> #include <string> #include<queue> #include<vector> #include<set> #include<map> #define LL long long using namespace std; const int N = 5005; LL n; LL m; string s; LL dp[N]; int go(int l, int r, int m, int k) { for (; l <= r; l++) if (dp[k] - dp[l] <= m) return l; return 0; } int bins(int l, int r, int m,int k) { while (l <= r) { if (r - l <= 5) return go(l, r, m,k); int mid = (l + r) / 2; if (dp[k] - dp[mid] <= m) r = mid; else l = mid + 1; } return 0; } int main() { cin.sync_with_stdio(false); //vector<LL> su; int t; cin >> t; while (t--) { cin >> m; cin >> s; int ans = 0; for (int i = 0; i < s.length(); i++) { int dif = 0,mxl=0,mxd=0; dp[0] = 0; for (int k = 1; i - k + 1 >= 0 && i + k < s.length(); k++) { int l = i - k + 1, r = i + k; dif += abs(s[l] - s[r]); dp[k] = dif; mxd = dif; mxl = k; //int len = bins(0, k, m,k); //ans = max(ans, k-len); } int st = 0, en = 1; while (en <= mxl) { if (dp[en]-dp[st] <= m) ans = max(ans, en - st),en++; else st++; } dif = 0, mxl = 0, mxd = 0; dp[0] = 0; for (int k = 1; i - k>= 0 && i + k < s.length(); k++) { int l = i - k, r = i + k; dif += abs(s[l] - s[r]); dp[k] = dif; mxd = dif; mxl = k; //int len = bins(0, k, m, k); //ans = max(ans, k - len); } st = 0, en = 1; while (en <= mxl) { if (dp[en] - dp[st] <= m) ans = max(ans, en - st), en++; else st++; } } cout << ans << endl; } return 0; }