I Count Two Three

二分查找用lower_bound

这道题用cin，cout会超时。。。

AC代码；

 1 /* */
 2 # include <iostream>
 3 # include <cstring>
 4 # include <string>
 5 # include <cstdio>
 6 # include <cmath>
 7 # include <algorithm>
 8 using namespace std;
 9
10 const int TWO = 2;
11 const int THREE = 3;
12 const int FIVE = 5;
13 const int SEVEN = 7;
14
15 const long long MAXN = 1e9+100;
16 const int N=1e6;
17 long long a[N];
18 int m;
19
20 void maketable()
21 {
22     m=0;
23     for(long long i=1; i<MAXN; i*=TWO )
24     {
25         for(long long j=1; j*i<MAXN; j*=THREE )
26         {
27             for(long long k=1; i*j*k<MAXN; k*=FIVE )
28             {
29                 for(long long l=1; i*j*k*l<MAXN; l*=SEVEN )
30                 {
31                     a[m++] = i*j*k*l;
32                 }
33             }
34         }
35     }
36 }
37
38 int main()
39 {
40     ios::sync_with_stdio(false);
41     maketable();
42     sort(a,a+m);
43     int t;
44     long long n;
45     scanf("%d", &t);
46     //cin>>t;
47     while( t-- )
48     {
49         scanf("%lld",&n);
50        printf("%lld\n", *lower_bound(a, a+m, n));
51 //        cin>>n;
52 //        cout<<*lower_bound(a,a+m,n)<<endl;
53     }
54     return 0;
55 }

原文地址：https://www.cnblogs.com/wsy107316/p/11462628.html