poj 1019 Number Sequence 【组合数学+数字x的位宽函数】

题目地址：http://poj.org/problem?id=1019

Number Sequence

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分析：序列如上所示，要求这个序列的第n位是什么，首先需要知道的是：一个数x的宽度怎么算？可以每次让x除10，看看多少次可以除尽。不过比较麻烦，有简单的算法：f[x]=log10(x)+1;  999的位宽=log10(999)+1=3;有了这个就可以来想怎么解决上面的问题。参考博客1：http://www.cnblogs.com/ACShiryu/archive/2011/08/05/2129009.html参考博客2：http://blog.csdn.net/lyy289065406/article/details/6648504一个大序列可以划分成许多个有规律的子序列，先找到第n位在那个子序列上，再找到在子序列的那个数字上，再找到在该数字的哪一位上。代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#include <cmath>
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define N 100000+100

using namespace std;
long long int f[40000];
long long int s[40000];
int main()
{
    int i, j;

    memset(f, 0, sizeof(f));
    memset(s, 0, sizeof(s));

    long long int sum=0;
    f[1]=1; s[1]=1;
    i=2;
    while( i<=31269 )
    {
        f[i]=f[i-1]+(int)log10((double)i)+1;
        s[i]=s[i-1]+f[i];
        i++;
    }//打表

    int tg; scanf("%d", &tg);
    while(tg--){
        int n;
        scanf("%d", &n);
        for(i=1; i<=31269; i++){
            if(s[i]>=n) break;
        }
        int pos=i; //找到在第pos个子序列上
        n=n-s[pos-1];
        for(j=1; j<=pos; j++){
            n=n-((int)log10((double)j)+1); //减去当前这个数的宽度
            if(n<=0){
                break;
            }
        }
        int pos2=j;//找到在子序列的第pos2个数上
        n=n+(int)log10((double)pos2)+1;//在这个数的第n位上

        int dd=(int)log10((double)pos2)+1-n;
        int q=1, w=1;
        for(i=1; i<=dd; i++) {q=q*10; w=w*10;} w=w*10;

        int ans;
        ans=pos2%w/q;
        printf("%d\n", ans );//输出结果
    }
    return 0;
}