poj 1019 Number Sequence 第i位上的数字 (组合数学)

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 37699 Accepted: 10890

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

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define len(x)   ( log10(x*1.0)+1 )typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn= 31268   ;int a[maxn+10];void pre(){    for(int i=1;i<=maxn+1;i++)    {        a[i]=a[i-1]+len(i);    }}int work(ll x){    ll s=0;int p;    for(int i=0;i<=maxn;i++)//从0开始    {        s+=a[i];        if(s+a[i+1]>=x)        {            p=i;            break;        }    }    ll leave=x-s;    s=0;    for(int i=1;i<=p+1;i++)    {        s+=len(i);        if(s>=leave)        {            p=i;            break;        }    }   return p/(int)( pow(10.0,1.0*s-leave)+0.5)%10;}int main(){    pre();    ll x;    int T;scanf("%d",&T);    while(T--)    {        scanf("%lld",&x);        printf("%d\n",work(x));    }   return 0;}