求解~ - 惟愿。。。

求解~

Problem Description
Let's define the function f(n)=⌊√n⌋.
Bo wanted to know the minimum number y which satisfies fy(n)=1.
note:f1(n)=f(n),fy(n)=f(fy−1(n))
It is a pity that Bo can only use 1 unit of time to calculate this function each time.
And Bo is impatient, he cannot stand waiting for longer than 5 units of time.
So Bo wants to know if he can solve this problem in 5 units of time.
 
Input
This problem has multi test cases(no more than 120).
Each test case contains a non-negative integer n(n<10100).
 
Output
For each test case print a integer - the answer y or a string "TAT" - Bo can't solve thisproblem.

 
Sample Input
233
233333333333333333333333333333333333333333333333333333333
 

Sample Output
3
TAT

AC代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<time.h>
 7 using namespace std;
 8 
 9 const int maxn = 2005;
10 const int N=100007;
11 
12 #define lson rt<<1///rt*2
13 #define rson rt<<1|1///rt*2+1
14 
15 char s[N];
16 
17 int main()
18 {
19     int len,i;
20 
21     while(scanf("%s", s) != EOF)
22     {
23         long long d=0;
24         len=strlen(s);
25         if(len>10)
26         {
27             printf("TAT\n");
28             continue ;
29         }
30         for(i=0; i<len; i++)
31             d=d*10+s[i]-'0';
32 
33 
34         if(d>=65536&&d<4294967296)
35             printf("5\n");
36         else if(d>=256&&d<65536)
37             printf("4\n");
38         else if(d>=16&&d<256)
39             printf("3\n");
40         else if(d>=4&&d<16)
41             printf("2\n");
42         else if(d>1&&d<4)
43             printf("1\n");
44         else if(d==1)
45             printf("0\n");
46         else
47             printf("TAT\n");
48     }
49     return 0;
50 }

Wa代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<time.h>
 7 using namespace std;
 8 
 9 const int maxn = 2005;
10 const int N=1000007;
11 
12 #define lson rt<<1///rt*2
13 #define rson rt<<1|1///rt*2+1
14 
15 char s[N];
16 
17 int main()
18 {
19     int len,i;
20 
21     while(scanf("%s", s) != EOF)
22     {
23         long long d=0;
24         len=strlen(s);
25         if(len>10)
26         {
27             printf("TAT\n");
28             continue ;
29         }
30         for(i=0; i<len; i++)
31             d=d*10+s[i]-'0';
32 
33         if(d<4294967296)
34         {
35             if(d>=65536)
36                 printf("5\n");
37             else if(d>=256)
38                 printf("4\n");
39             else if(d>=16)
40                 printf("3\n");
41             else if(d>=4)
42                 printf("2\n");
43             else if(d>1)
44                 printf("1\n");
45             else if(d==1)
46                 printf("0\n");
47         }
48         else
49             printf("TAT\n");
50     }
51     return 0;
52 }

很搞不懂，谁来解释一下 if else 1号和 if else2号的区别，~~~~(>_<)~~~~

posted on 2016-07-26 14:00 惟愿。。。阅读(163) 评论(0) 编辑收藏举报