POJ 1019 Number Sequence 难

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Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32214		Accepted: 9197

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

2
8
3

Sample Output

2
2

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

给你一串数字，让你找第n个数字是几。

参考链接：点击打开链接

//904K	0MS
#include<stdio.h>
#include<math.h>
#define M 31269
long long a[M],s[M];
void init()
{
    a[1]=s[1]=1;
    for(int i=2;i<M;i++)
    {
        a[i]=a[i-1]+(int)(log10((double)i)+1);
        s[i]=s[i-1]+a[i];
    }
}
int solve(int x)
{
    int i=1;
    while(s[i]<x)
        i++;
    int pos=x-s[i-1];
    int len=0;
    for(i=1;len<pos;i++)
        len+=(int)log10(i)+1;
    return (i-1)/(int)pow((double)10,(len-pos))%10;
}
int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        printf("%d\n",solve(n));
    }
    return 0;
}