Find the nth digit

Find the nth digitTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) Total Submission (s): 8801 Accepted Submission (s): 2504

problem DescriptionAssumptions:
S1 = 1
S2 = 12
S3 = 123
S4 = 1234
.........
S9 = 123456789
S10 = 1234567891
S11 = 12345678912
............
S18 = 123456789123456789
..................
Now we're going to connect all the strings together.
S = 1121231234.......123456789123456789112345678912 .....
So can you tell me what the nth number is in the s string?

Inputthe input is first a number k, which represents a K-time inquiry.
The next K line has an integer n (1 <= n < 2^31) per line.
Outputfor each n, the nth corresponding number in the output s.

Sample Input

61234510

Sample Output

112124Problem Solving Ideas:First find the nth number in Si,This is achieved by the formula I (i+1)/2=n i= (2n+0.25) 0.5-0.5, where I should be rounded up instead of rounding down. Then the nth number is the number of n-si-1 in Si, and then mod 9 can be obtained results for data above S8 directly by checking the tableSource code:#include <stdio.h> #include <math.h> #include <stdlib.h>int main () {int k;    Double i;    unsigned long j,n;    int s[]={0,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,1,2,3,4,5,6,7, 1,2,3,4,5,6,7,8};    scanf ("%d", &k);      while (k--) {scanf ("%ld", &n);      if (n<=36) printf ("%d\n", S[n]);        else {I=ceil (sqrt (2*n*1.0+0.25)-0.5);        j= (unsigned long) i-1; n=n-j* (j+1)/2;// printf ("i=%lf\n j=%ld\n n1=%ld\n n2=%ld\n", i,j,j* (j+1)/2,n);        if (n%9) printf ("%d\n", n%9);       else printf ("9\n");    }} system ("Pause");    return 0; }

Find the nth digit