find the nth digit
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5514 Accepted Submission(s): 1557
Problem Description
假设:
S1 = 1
S2 = 12
S3 = 123
S4 = 1234
.........
S9 = 123456789
S10 = 1234567891
S11 = 12345678912
............
S18 = 123456789123456789
..................
现在我们把所有的串连接起来
S = 1121231234.......123456789123456789112345678912.........
那么你能告诉我在S串中的第N个数字是多少吗?
S1 = 1
S2 = 12
S3 = 123
S4 = 1234
.........
S9 = 123456789
S10 = 1234567891
S11 = 12345678912
............
S18 = 123456789123456789
..................
现在我们把所有的串连接起来
S = 1121231234.......123456789123456789112345678912.........
那么你能告诉我在S串中的第N个数字是多少吗?
Input
输入首先是一个数字K,代表有K次询问。
接下来的K行每行有一个整数N(1 <= N < 2^31)。
接下来的K行每行有一个整数N(1 <= N < 2^31)。
Output
对于每个N,输出S中第N个对应的数字.
Sample Input
6 1 2 3 4 5 10
Sample Output
1 1 2 1 2 4
Author
8600
Source
Recommend
8600
注意一下数据类型要用__int64
code:
1 #include <iostream> 2 #include <iomanip> 3 #include <fstream> 4 #include <sstream> 5 #include <algorithm> 6 #include <string> 7 #include <set> 8 #include <utility> 9 #include <queue> 10 #include <stack> 11 #include <list> 12 #include <vector> 13 #include <cstdio> 14 #include <cstdlib> 15 #include <cstring> 16 #include <cmath> 17 #include <ctime> 18 #include <ctype.h> 19 using namespace std; 20 21 int main() 22 { 23 __int64 k; 24 __int64 i; 25 __int64 temp; 26 __int64 n; 27 while(~scanf("%I64d",&k)) 28 { 29 while(k--) 30 { 31 scanf("%I64d",&n); 32 temp=(-1+sqrt(double(1+8*(n-1))))/2; 33 if(temp%2) 34 n-=((1+temp)/2)*temp; 35 else 36 n-=(1+temp)*(temp/2); 37 n%=9; 38 if(n==0) 39 n=9; 40 printf("%I64d\n",n); 41 } 42 } 43 return 0; 44 }