1. The problem is to
write two retrieval algorithms: find x in a sorted array T[1...n], if x is in T, output x in the subscript j of T; if x is not in T, output j =0. According to the experiment template, only the complexity results are given in the "analysis" part.
2. Analysis
1.
Store the array data through a tag array, the tag array subscript is the data value, and the value is the subscript corresponding to the original data to achieve O(1) query;
2.
Binary search, if found, return the subscript, if not found, it will be 0;
3. Design
1.
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<iomanip>
#include<map>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<set>
#include<cctype>
#include<string>
#include<stdexcept>
#include<fstream>
#include<sstream>
#include<sstream>
#define mem(a,b) memset(a,b,sizeof(a))
#define debug() puts("what the fuck!")
#define dedebug() puts("what the fuck!!!")
#define ll long long
#define ull unsigned long long
#define speed {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); };
using namespace std;
const double PI = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 2e2 + 10;
const ll INF = 1e18;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double esp_0 = 1e-6;
const double gold = (1 + sqrt(5)) / 2;
template<typename T>
inline void rd(T&x) {
int temp = 1;
char c = getchar();
x = 0;
while (c > '9' || c < '0') {
if (c == '-')temp = -1; c = getchar();}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';c = getchar();}
x *= temp;
}
ll gcd(ll x, ll y) {
return y ? gcd(y, x % y) : x;}
int vis[maxn];
int n;
signed main() {
cin >> n;
mem(vis, 0);
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
vis[x] = i;
}
int q;
while (cin >> q) {
if (vis[q])cout << vis[q] << endl;
else cout << 0 << endl;
}
return 0;
}
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<iomanip>
#include<map>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<set>
#include<cctype>
#include<string>
#include<stdexcept>
#include<fstream>
#include<sstream>
#include<sstream>
#define mem(a,b) memset(a,b,sizeof(a))
#define debug() puts("what the fuck!")
#define dedebug() puts("what the fuck!!!")
#define ll long long
#define ull unsigned long long
#define speed {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); };
using namespace std;
const double PI = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 2e2 + 10;
const ll INF = 1e18;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double esp_0 = 1e-6;
const double gold = (1 + sqrt(5)) / 2;
template<typename T>
inline void rd(T&x) {
int temp = 1;
char c = getchar();
x = 0;
while (c > '9' || c < '0') {
if (c == '-')temp = -1; c = getchar();}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';c = getchar();}
x *= temp;
}
ll gcd(ll x, ll y) {
return y ? gcd(y, x % y) : x;}
int a[maxn];
int vis[maxn];
int n;
signed main() {
cin >> n;
mem(vis, 0);
for (int i = 1; i <= n; ++i) {
cin >> a[i];
vis[a[i]] = i;
}
sort(a + 1, a + 1 + n);
int q;
while (cin >> q) {
int pos = lower_bound(a + 1, a + 1 + n, q) - a;
if (a[pos] ==q)cout << vis[a[pos]] << endl;
else cout << 0 << endl;
}
return 0;
}
4. Analysis
- O （1）
- Log in
5. Source code