题面
说几个坑点，一定要看提示，要不然题目感觉不是很能理解出题人的意思。
看了提示要注意到时 \(a_i\) 的倍数，而不是所有 \(a\) 的最大值。
并要注意到他按顺序排列也是要遵守对齐规则的。
思路
思路很简单，就是把所有结构体存下来，然后按照他给的公式算大小和对齐值。
然后再加入的时候先把他放进去，等后面再来一个元素时，根据公式算出这个元素的起始位置，然后减去上一个地址，得到上一个元素的大小。
在查询元素的起始位置时，就暴力的每次进去一维算一下前缀和即可。
在查询位置的元素时，就是算前缀和和当前位置比大小就好了。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
vector<vector<string>> g, gx; // 名字 类型名
vector<vector<ll>> len; // 长度
map<string, ll> mp, dyc, lenn;
ll inn;
string nm[105];
string lxnm[105];
ll dx[105];
int ntt;
ll nadd, lstlen;
signed main()
{ios::sync_with_stdio(0), cin.tie(0);int _;cin >> _;dyc["byte"] = 1;dyc["short"] = 2;dyc["int"] = 4;dyc["long"] = 8;lenn["byte"] = 1;lenn["short"] = 2;lenn["int"] = 4;lenn["long"] = 8;while (_--){int op;cin >> op;// cout << op << " ";if (op == 1){string lxm;cin >> lxm;ll k;cin >> k;mp[lxm] = ntt++;vector<string> mz, lx;vector<ll> lz;ll mx = 0, tlen = 0;ll tot = 0, lst = 0;for (int i = 1; i <= k; i++){string lm, s;cin >> lm >> s;if (i > 1){tot = (tot + lst + dyc[lm] - 1) / dyc[lm] * dyc[lm];}mz.push_back(s);lz.push_back(lenn[lm]);lx.push_back(lm);tlen += lenn[lm];mx = max(mx, dyc[lm]);lst = lenn[lm];}g.push_back(mz);gx.push_back(lx);len.push_back(lz);dyc[lxm] = mx;lenn[lxm] = (tot + lst + mx - 1) / mx * mx;cout << lenn[lxm] << " " << mx << "\n";}else if (op == 2){cin >> lxnm[inn];cin >> nm[inn];ll le = lenn[lxnm[inn]];ll lstadd = nadd;nadd -= lstlen;if (inn){nadd = (nadd + lstlen + dyc[lxnm[inn]] - 1) / dyc[lxnm[inn]] * dyc[lxnm[inn]];dx[inn] = le;dx[inn - 1] += nadd - lstadd;}else{dx[inn] = le;}cout << nadd << "\n";nadd += le;lstlen = le;inn++;}else if (op == 3){string s;cin >> s;int i = 0;string nw = "";while (i < s.size() && s[i] != '.'){nw += s[i];i++;}i++;string lxm;ll ans = 0;for (int j = 0; j < inn; j++){if (nm[j] == nw){lxm = lxnm[j];break;}ans += dx[j];}for (; i < s.size(); i++){ll nmx = dyc[lxm];ll nid = mp[lxm];string nw = "";while (i < s.size() && s[i] != '.'){nw += s[i];i++;}ll tot = 0, lst = 0;for (int j = 0; j < g[nid].size(); j++){if (j){tot = (tot + lst + dyc[gx[nid][j]] - 1) / dyc[gx[nid][j]] * dyc[gx[nid][j]];}if (g[nid][j] == nw){ans += tot;lxm = gx[nid][j];break;}lst = lenn[gx[nid][j]];}// cout << nw << " " << ans << "\n";}cout << ans << "\n";}else{ll x;cin >> x;if (x >= nadd){cout << "ERR\n";continue;}string ans = "";string lxm;for (int j = 0; j < inn; j++){// cout << dx[j] << "\n";if (x < dx[j]){lxm = lxnm[j];ans += nm[j];break;}x -= dx[j];}while (1){if (lxm == "byte" || lxm == "int" || lxm == "short" || lxm == "long"){if (x >= dyc[lxm]){ans = "ERR";}break;}// cout << x << "\n";ll nmx = dyc[lxm];ll nid = mp[lxm];bool flg = 0;ll tot = 0, lst = 0, lsttot = 0;for (int j = 0; j < g[nid].size(); j++){if (j){tot = (tot + lst + dyc[gx[nid][j]] - 1) / dyc[gx[nid][j]] * dyc[gx[nid][j]];}// cout << "*" << tot << "\n";if (x < tot){lxm = gx[nid][j - 1];ans += ".";ans += g[nid][j - 1];flg = 1;x -= lsttot;break;}lst = lenn[gx[nid][j]];lsttot = tot;}if (!flg){tot = (tot + lst + dyc[lxm] - 1) / dyc[lxm] * dyc[lxm];// cout << "*" << tot << "\n";if (x <= tot){lxm = gx[nid][g[nid].size() - 1];ans += ".";ans += g[nid][g[nid].size() - 1];flg = 1;x -= lsttot;}}// cout << "*" << ans << "\n";if (!flg){ans = "ERR";break;}}cout << ans << "\n";}}return 0;
}总结
先好好读题（读完，说不定出题人写了提示呢）。
注意每道题的边界情况，别把代码写出来后再去想边界。