题目链接https://loj.ac/problem/6282
大意是让一个数插入某个数之前，然后在让你查询某点的值
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 const int maxn = 1e5 + 10; 6 const int inf = 0x3f3f3f3f; 7 int dir[10][10] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; 8 int a[maxn], b[maxn]; 9 int opt, l, r, c, m, t, n; 10 vector<int > v[1010]; 11 struct st 12 { 13 int x, y; 14 }; 15 16 inline int read() 17 { 18 char ch = getchar(); ll k = 0, f = 1; 19 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 20 while(ch >= '0' && ch <= '9') {k = (k << 1) + (k << 3) + ch - '0'; ch = getchar();} 21 return k * f; 22 } 23 24 inline st findy(int x) 25 { 26 int i = 1; 27 while(x > v[i].size()) 28 x -= v[i].size(), ++i; 29 return (st) {i, x - 1} ; 30 } 31 32 inline void rebuild() // 重新分块 33 { 34 int cnt = 0; 35 for(int i = 1; i <= m; ++i) 36 { 37 int s = v[i].size(); 38 for(int j = 0; j <= s - 1; ++j) 39 { 40 b[++cnt] = v[i][j]; 41 } 42 v[i].clear(); 43 } 44 t = sqrt(cnt), m = (cnt - 1) / t + 1; 45 for(int i = 1; i <= cnt; ++i) 46 { 47 v[(i - 1) / t + 1].push_back(b[i]); 48 } 49 } 50 inline void inser(int l, int r) 51 { 52 st k = findy(l); 53 int x = k.x, y = k.y; 54 v[x].insert(v[x].begin() + y, r); 55 if(v[x].size() > 10 * t) rebuild(); //体现分块的思想，如果一直往一个块插入数据那不和暴力没区别了吗，因此块内数据过多时要重新分块。 56 } 57 58 int main() 59 { 60 std::ios::sync_with_stdio(false);std::cin.tie(0); 61 n = read(); 62 for(int i = 1; i <= n; ++i) 63 { 64 a[i] = read(); 65 } 66 67 t = sqrt(n); m = (n - 1) / t + 1; 68 for(int i = 1; i <= n; ++i) 69 { 70 v[(i - 1) / t + 1].push_back(a[i]); 71 } 72 73 for(int i = 1; i <= n; ++i) 74 { 75 opt = read(); l = read(); r = read(); c = read(); 76 if(opt == 0) inser(l, r); 77 else 78 { 79 st ch = findy(r); 80 cout<< v[ch.x][ch.y] <<endl; 81 } 82 } 83 return 0; 84 }
这题数据很水，直接vector insert暴力就能过，当然我不建议。。