C - Number Sequence
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
2 8 3
Sample Output
2 2
#include<cstdio>
#include<stack>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=225;
const int mmax = 40200+ 7;
long long len[mmax],sum[mmax];
void init() //打表
{
len[0]=0;
sum[0]=0;
//分段 1 12 123 1234 …… 123456891011
for(int i=1;i<mmax;i++){
len[i]=len[i-1]+(int)log10((double)i)+1; //第i段
sum[i]=sum[i-1]+len[i]; //加入第i段后的整个长度
}
}
int solve(int n){
int k=0;
while(sum[k]<n) //能覆盖n的最长序列
k++;
int lastn=n-sum[k-1]; //去掉完全段后的剩余部分
int len_n=0;
int t=1;
while(len_n<lastn){ //求残余部分在第k段的位置
len_n+=(int)log10((double)t)+1;
t++;
}
int num=len_n-lastn;
return (t-1)/(int)pow((double)10,(double)num)%10;
//pow原型为:double pow(double x, double y);
}
int main()
{
init();
int t;
cin>>t;
int n;
while(t--){
scanf("%d",&n);
printf("%d\n",solve(n));
}
return 0;
}