Number Sequence
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 35823 | Accepted: 10340 |
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
题目大意：找出第i位的数值
思路：可以把序列分成若干组如1 12 123 1234 ....利用s[]a[]两个数组分别记录全部每序列的长度，和当前每组序列的长度，然后大体的找到i在哪一组，然后再在每一个
小的组里找到相应位置的这个值。
#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,j,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("%lld",&cla);
while(cla--)
{
scanf("%lld",&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("%lld\n",so(i,a[i]-m) );//<span style="font-family: Consolas, 'Courier New', Courier, mono, serif; line-height: 18px; background-color: rgb(248, 248, 248);">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</span>
}
return 0;
}
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