poj 1019

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33838 Accepted: 9679

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

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

这一题的关键是要想得到对于数字n，它的位数是多少。及len(n)=log(10)n + 1。也就是12是2位数，182是三位数，然后剩下就就好办了，相信一下的代码应该很好理解了。

AC代码：

#include<iostream>#include<cmath>using namespace std;long long a[31270],s[31270];void init(){    a[0]=s[0]=0;    a[1]=s[1]=1;    for(int i=2;i<31270;i++){        a[i]=a[i-1]+(int)log10(i+0.0)+1;        s[i]=s[i-1]+a[i];    }}int main(){    init();    int T; cin>>T;    while(T--){        long long n; cin>>n;        int i; i=1;        int len,pos;        while(s[i]<n)            i++;        pos=n-s[i-1];        len=0;        for(i=1;len<pos;i++)            len+=log10(i+0.0)+1;        int ans=(i-1)/pow(10.0,len-pos+0.0);        ans%=10;        cout<<ans<<endl;    }    return 0;}