[hdu5204]水题
思路：插入的数按指数级增长，所以范围内最多存在logR个数。并且最近i次插入的数，首位置为2^(i-1)，且每隔2^i出现一次，于是暴力之。。可以用插入排序维护，也可查询时再排下序。
一：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define lson l, m, rt << 1 26 #define rson m + 1, r, rt << 1 | 1 27 #define define_m int m = (l + r) >> 1 28 #define rep0(a, b) for (int a = 0; a < (b); a++) 29 #define rep1(a, b) for (int a = 1; a <= (b); a++) 30 #define all(a) (a).begin(), (a).end() 31 #define lowbit(x) ((x) & (-(x))) 32 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 33 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 34 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 35 #define pchr(a) putchar(a) 36 #define pstr(a) printf("%s", a) 37 #define sint(a) ReadInt(a) 38 #define sint2(a, b) ReadInt(a);ReadInt(b) 39 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c) 40 #define pint(a) WriteInt(a) 41 42 typedef double db; 43 typedef long long LL; 44 typedef pair<int, int> pii; 45 typedef multiset<int> msi; 46 typedef set<int> si; 47 typedef vector<int> vi; 48 typedef map<int, int> mii; 49 50 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1}; 51 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1}; 52 const int maxn = 1e3 + 7; 53 const int maxm = 1e5 + 7; 54 const int maxv = 1e7 + 7; 55 const int max_val = 1e6 + 7; 56 const int MD = 1e9 +7; 57 const int INF = 1e9 + 7; 58 const double PI = acos(-1.0); 59 const double eps = 1e-10; 60 61 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 62 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}} 63 template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);} 64 65 66 struct abc { 67 pii a[100007]; 68 int l, r; 69 void Init() { l = r = 0; } 70 void push_back(int x) { 71 a[r++] = make_pair(x, 0); 72 for(int i = l; i < r - 1; i++) a[i].second++; 73 if (r - l >= 62) { 74 int pos; 75 for (int i = l; i < r; i++) { 76 if (a[i].second == 61) { 77 pos = i; 78 break; 79 } 80 } 81 for (int i = pos; i > l; i--) a[i] = a[i - 1]; 82 l++; 83 } 84 int p = r - 1; 85 while (p > l && a[p].first < a[p - 1].first) { 86 swap(a[p], a[p - 1]); 87 p--; 88 } 89 } 90 pii &operator [] (int i) { 91 return a[l + i]; 92 } 93 int size() { 94 return r - l; 95 } 96 }; 97 98 99 abc g; 100 101 LL calc(LL x, LL pos) { 102 if (x < pos) return 0; 103 return (x - pos) / pos / 2 + 1; 104 } 105 int main() { 106 //freopen("in.txt", "r", stdin); 107 int n; 108 while (cin >> n) { 109 g.Init(); 110 rep0(i, n) { 111 int id, w; 112 sint(id); 113 if (id == 1) { 114 sint(w); 115 g.push_back(w); 116 } 117 else { 118 LL L, R, k; 119 sint3(L, R, k); 120 int sz = g.size(); 121 rep0(i, sz) { 122 LL pos = 1LL << g[i].second, c = calc(R, pos) - calc(L - 1, pos); 123 if (k <= c) { 124 pint(g[i].first); 125 pchr('\n'); 126 break; 127 } 128 k -= c; 129 } 130 } 131 } 132 } 133 return 0; 134 }
二：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define rep0(a, b) for (int a = 0; a < (b); a++) 24 #define rep1(a, b) for (int a = 1; a <= (b); a++) 25 #define all(a) (a).begin(), (a).end() 26 #define lowbit(x) ((x) & (-(x))) 27 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 28 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 29 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 30 #define pchr(a) putchar(a) 31 #define pstr(a) printf("%s", a) 32 #define sint(a) ReadInt(a) 33 #define sint2(a, b) ReadInt(a);ReadInt(b) 34 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c) 35 #define pint(a) WriteInt(a) 36 37 typedef double db; 38 typedef long long LL; 39 typedef pair<int, int> pii; 40 typedef multiset<int> msi; 41 typedef set<int> si; 42 typedef vector<int> vi; 43 typedef map<int, int> mii; 44 45 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1}; 46 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1}; 47 const int maxn = 1e3 + 7; 48 const int maxm = 1e5 + 7; 49 const int maxv = 1e7 + 7; 50 const int max_val = 1e6 + 7; 51 const int MD = 1e9 +7; 52 const int INF = 1e9 + 7; 53 const double PI = acos(-1.0); 54 const double eps = 1e-10; 55 56 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 57 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}} 58 template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);} 59 60 struct abc { 61 int a[110000]; 62 int l, r; 63 void Init() { l = r = 0; } 64 void push_back(int x) { 65 a[r++] = x; 66 if (r - l >= 62) { 67 l++; 68 } 69 } 70 int &operator [] (int i) { 71 return a[l + i]; 72 } 73 int size() { 74 return r - l; 75 } 76 }; 77 78 abc g; 79 80 pair<int, LL> a[100]; 81 82 LL calc(LL x, int id) { 83 LL start = 1LL << (g.size() - id - 1), t = 1LL << (g.size() - id); 84 if (x < start) return 0; 85 return (x - start) / t + 1; 86 } 87 88 int main() { 89 //freopen("in.txt", "r", stdin); 90 int n; 91 while (cin >> n) { 92 g.Init(); 93 rep0(i, n) { 94 int id, w; 95 sint(id); 96 if (id == 1) { 97 sint(w); 98 g.push_back(w); 99 } 100 else { 101 LL L, R, k; 102 sint3(L, R, k); 103 int total = 0, sz = g.size(); 104 rep0(i, sz) { 105 LL c = calc(R, i) - calc(L - 1, i); 106 if (c > 0) a[total++] = make_pair(g[i], c); 107 } 108 sort(a, a + total); 109 int now = 0; 110 while (1) { 111 if (k <= a[now].second) { 112 break; 113 } 114 k -= a[now++].second; 115 } 116 pint(a[now].first); 117 pchr('\n'); 118 } 119 } 120 } 121 return 0; 122 }
