POJ1019 Number Sequence

Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36256		Accepted: 10461

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

 

题意就不说了，看下就明白。思路就是打表，然后就是要知道求一个数的位数：(int)log10((double)x)+1

 

/*
ID: LinKArftc
PROG: 1019.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 40000;

ll sum[maxn], line[maxn];//分别是前i行的位数和，第i行的位数

int getbit(int x) {
    return (int)log10((double)x) + 1;
}

void init() {
    sum[1] = line[1] = 1;
    for (int i = 2; i <= 35000; i ++) {
        line[i] = line[i-1] + getbit(i);
        sum[i] = sum[i-1] + line[i];
    }
}

int getpos(int x, int pos) {
    int len = getbit(x);
    for (int i = 1; i <= len - pos; i ++) x /= 10;
    return x % 10;
}

int main() {

    init();
    int T;
    ll n;
    scanf("%d", &T);
    while (T --) {
        scanf("%lld", &n);
        int cur = 1;
        while (sum[cur] < n) cur ++;
        ll pos = n - sum[cur-1];
        while (cur >= 1) {
            if (pos > line[cur-1]) {
                pos -= line[cur-1];
                printf("%d
", getpos(cur, pos));
                break;
            } else cur --;
        }
    }

    return 0;
}