Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second

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…S_k. Each group S_k consists of a sequence of positive integer numbers ranging from 1 to k, written one after another. For example, the first 80digits of the sequence are as follows:

11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 <=t <=25), 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                           Output for Sample Input

2 8 3

2 2

题意： 给定一个数字n，要求出第n位对应题目给定那个序列的数字是多少，序列为1 ， 1， 2， 1， 2， 3， 1， 2， 3， 4.。。。以此类推。注意2位数是占2个数，3位数是占3个数这样的。

思路:题目数据为2^31 - 1。直接暴力肯定跪。所以可以分区间讨论，个位数，十位数，百位数，这样去讨论，个位数有9个区间1,[1,2],[1,3]...[1,9]一共45位。以此类推。可以位数对应的第几个区间。打出一个表来。然后根据位数在区间去查找。最后找出对应位数上的数字。

代码：

#include <stdio.h>

#include <string.h>

#include <math.h>

int t, n, pos;

long long num[100005];

void init() {//我去年打了个表

    pos = 0;

    memset(num, 0, sizeof(num));

    for (int i = 1; i <= 5; i ++) {

	for (long long j = pow(10, i - 1); j < pow(10, i); j ++) {

	    num[j] = num[j - 1];

	    int d = (j - pow(10, i - 1) + 1) * i;

	    for (long long k = 1; k < i; k ++)

		d += (pow(10, k) - pow(10, k - 1)) * k;

	    num[j] += d;

	}

    }

}

void find() {//找出对应区间

    for (int i = 1; i < 100005; i ++)

	if (n > num[i - 1] && n <= num[i]) {

	    pos = i;

	    break;

	}

}

int solve() {//找出对应数字。

    int wei = n - num[pos - 1];

    int sum = 0;

    for (int i = 1; i <= pos; i ++) {

	int sb = i;

	int j;

	for (j = 5; j >= 0; j --)

	    if (sb >= int(pow(10, j)))

		break;

	for (; j >= 0; j --) {

	    if (sb >= int(pow(10, j))) {

		int fuck = sb / int(pow(10, j));

		sb -= fuck * int(pow(10, j));

		sum ++;

		if (sum == wei)

		    return fuck;

	    }

	    else {

		sum ++;

		if (sum == wei)

		    return 0;

	    }

	}

    }

}

int main() {

    init();

    scanf("%d", &t);

    while (t --) {

	scanf("%d", &n);

	find();

	printf("%d\n", solve());

    }

    return 0;

}