O - Number Sequence

Number Sequence

题目描述

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

Iutput

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

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思路：

模拟分组，把 1 看做第 1 组，12 看做第 2 组，123 看做第 3 组……那么第 i 组就是存放数字序列为 [ 1，i ] 的正整数，但第 i 组的长度不一定是 i 已知输入查找第 n 个位的 n 的范围为 (1 ≤ n ≤ 2147483647 )，那么至少要有 31268 个组才能使得数字序列达到有第 2147483647 位 注意：2147483647 刚好是 int 的正整数最大极限值( )，所以对于 n 用 int 定义就足矣。但是 pos [ 31268 ] 存在超过 2147483647 的位数，因此要用 unsigned 或 long long 之类的去定义 pos [ ]

其中：( int ) log10 ( i * 1.0 ) + 1 是求某个数所占的空间

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代码

#include<iostream> #include<cstdio> #include<string.h> #include<string> #include<cmath> #include<queue> #define LL long long using namespace std; LL s[100000],a[100000]; LL so(LL x,LL d) { while(d--) { x/=10; } return x%10; } int main() { LL n,m,i,cla; a[1]=s[1]=1; for(i=2;i<=34000;i++)//先打表 { a[i]=a[i-1]+(LL)log10((double)i)+1; s[i]=s[i-1]+a[i]; } scanf("%I64d",&cla); while(cla--) { scanf("%I64d",&n); i=0; while(s[i]<n) i++; m=n-s[i-1];//找到在n在全部序列里的下标 i=0; while(m>a[i]) i++;//i为所在n组的长度 printf("%I64d\n",so(i,a[i]-m) ); //例如要取出1234的2，那么多余的位数有2位：34。 //那么用1234 / 100，得到12，再对12取模10，就得到2 } return 0; }