poj1019——log10求位数

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35084		Accepted: 10099

Description

A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another. 

For example, the first 80 digits of the sequence are as follows: 

11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 ≤ i ≤ 2147483647)

Output

There should be one output line per test case containing the digit located in the position i.

Sample Input

283

Sample Output

22 题意：求数串112123123412345..第n位的数字(0-9) 思路：取将每一个123..的位数存起来，利用取对数求位数公式：(int)log10(n*1.0)+1，先二分找出n所在的那个串的位数， n-=其前缀，再找到第n位即可      庆祝一下，程序完全是自己调试出来的，从WA到AC，这次终于没有借助题解了

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            using namespace std;typedef unsigned long long ull;const int maxn=10001000;const ull INF=21474836470;int T;int n;ull a[maxn],s[maxn];int cnt=1;int tag=1;ull mypow(int n,int k){ ull res=1; while(k--) res*=n; return res;}ull BinSearch(ull *a,int left,int right,int key){ while(left
            
             =key&&a[mid-1]
             
              =key) right=mid; else left=mid+1; //cout<
              
               <
               
                >T; while(T--){ cin>>n; //cout<<"ans="<
                
                 <