HDU 5752 Sqrt Bo

Sqrt Bo

Time limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 980 Accepted Submission (s): 452

Problem Descriptionlet ' s define the functionf(n)=⌊n√⌋ .

Bo wanted to know the minimum numberywhich satisfiesfy(n)=1 .

NoteF1(N)=F(N),fy(n)=F(fy−1(n))

It is a pity, so 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.

Inputthis problem have multi test cases (no more than).

Each test case contains a non-negative integer n(n<ten) /c12>.

Outputfor each test case print a integer-the answer y or a string "TAT"-Bo can ' t solve this problem.

Sample Input233233333333333333333333333333333333333333333333333333333333

Sample Output3tat

Author Shaoxing one middle

Source2016 multi-university Training Contest 3

#include <iostream>#include<stdio.h>#include<math.h>#include<string>#include<string.h>using namespacestd;intMain () {strings6="4294967296"; strings5="65536"; strings4=" the"; strings3=" -"; stringS2="4"; strings1="2"; stringA;  while(cin>>a) {if((A.length () ==s6.length () &&a>=s6) | | (A.length () >s6.length ())) {printf ("tat\n"); }        Else if((A.length () ==s5.length () &&a>=s5) | | (A.length () >s5.length ())) {printf ("5\n");} Else if((A.length () ==s4.length () &&a>=s4) | | (A.length () >s4.length ())) {printf ("4\n");} Else if((A.length () ==s3.length () &&a>=s3) | | (A.length () >s3.length ())) {printf ("3\n");} Else if((A.length () ==s2.length () &&a>=s2) | | (A.length () >s2.length ())) {printf ("2\n");} Else if((A.length () ==s1.length () &&a>=s1) | | (A.length () >s1.length ())) {printf ("1\n");} Else if(a=="1") printf ("0\n"); Else if(a=="0") printf ("tat\n"); }    return 0;}

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HDU 5752 Sqrt Bo