问题：
两种检索算法：在一个排好序的数组T[1..n]中查找x，如果x在T中，输出x在T的下标j；如果x不在T中，输出j=0.
解析：
①直接遍历数组，查找x。
②二分查找。
设计（核心代码）：
①遍历
1 void solve1(int x)
2 {
3 for(int i=1;i<=n;++i)
4 {
5 if(a[i]==x)
6 {
7 printf("%d
",i);
8 return ;
9 }
10 }
11 printf("0
");
12 return ;
13 }
②二分
1 void solve2(int x)
2 {
3 int l=1,r=n,mid;
4 while(l<=r)
5 {
6 mid=(l+r)>>1;
7 if(a[mid]>x)
8 r=mid-1;
9 else if(a[mid]<x)
10 l=mid+1;
11 else
12 {
13 printf("%d
",mid);
14 return ;
15 }
16 }
17 printf("0
");
18 return ;
19 }
分析：
①遍历数组，复杂度O(n).
②二分查找，复杂度O(logn).
源码：
https://github.com/Big-Kelly/Algorithm
1 #include<bits/stdc++.h>
2 #include <set>
3 #include <map>
4 #include <stack>
5 #include <cmath>
6 #include <queue>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13
14 #define ll long long
15 #define pll pair<ll,ll>
16 #define pii pair<int,int>
17 #define bug printf("*********
")
18 #define FIN freopen("input.txt","r",stdin);
19 #define FON freopen("output.txt","w+",stdout);
20 #define IO ios::sync_with_stdio(false),cin.tie(0)
21 #define ls root<<1
22 #define rs root<<1|1
23 #define Q(a) cout<<a<<endl
24
25 using namespace std;
26 const int inf = 2e9 + 7;
27 const ll Inf = 1e18 + 7;
28 const int maxn = 1e6 + 5;
29 const int mod = 1e9 + 7;
30
31 ll gcd(ll a, ll b)
32 {
33 return b ? gcd(b, a % b) : a;
34 }
35
36 ll lcm(ll a, ll b)
37 {
38 return a / gcd(a, b) * b;
39 }
40
41 ll read()
42 {
43 ll p = 0, sum = 0;
44 char ch;
45 ch = getchar();
46 while (1)
47 {
48 if (ch == '-' || (ch >= '0' && ch <= '9'))
49 break;
50 ch = getchar();
51 }
52
53 if (ch == '-')
54 {
55 p = 1;
56 ch = getchar();
57 }
58 while (ch >= '0' && ch <= '9')
59 {
60 sum = sum * 10 + ch - '0';
61 ch = getchar();
62 }
63 return p ? -sum : sum;
64 }
65
66 int a[maxn],n;
67
68 void solve1(int x)
69 {
70 for(int i=1;i<=n;++i)
71 {
72 if(a[i]==x)
73 {
74 printf("%d
",i);
75 return ;
76 }
77 }
78 printf("0
");
79 return ;
80 }
81
82 void solve2(int x)
83 {
84 int l=1,r=n,mid;
85 while(l<=r)
86 {
87 mid=(l+r)>>1;
88 if(a[mid]>x)
89 r=mid-1;
90 else if(a[mid]<x)
91 l=mid+1;
92 else
93 {
94 printf("%d
",mid);
95 return ;
96 }
97 }
98 printf("0
");
99 return ;
100 }
101
102 int main()
103 {
104 scanf("%d",&n);
105 for(int i=1;i<=n;++i)
106 scanf("%d",&a[i]);
107 sort(a+1,a+1+n);
108 int q;
109 scanf("%d",&q);
110 while(q--)
111 {
112 int x;
113 scanf("%d",&x);
114 solve1(x);
115 solve2(x);
116 }
117 }