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
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910
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 3Sample Output
2 2
题意：给出一串特殊的数字，输入t表示有多少组数据，在输入i，输出该数第i位为什么，i超大所以该数只能用数组存储，又因为其规律为1，12，123，1234，12345，所以其规律作为一个单位，用一个数组存储每个单位有几位，再用一个存储当前单位总共几位。不过有些地方不太理解
#include<iostream>
#include<cstdio>
#include<math.h>
using namespace std;
unsigned int num[31270],sum[31270];
void sequence();
int judge(int a);
int main()
{
int t,n;
sequence();
cin>>t;
while(t--)
{
cin>>n;
cout<<judge(n)<<endl;
}
}
void sequence()
{
int i,j,k;
num[1]=1,sum[1]=1;
for(i=2;i<31270;i++)
{
num[i]=num[i-1]+(int)log10((double)i)+1;//到10时，就会有两位，所以log10保证了位数的准确
sum[i]=sum[i-1]+num[i];//当前总位数
}
}
int judge(int a)
{
int i=1,digit,temp;
while(sum[i]<a)//找在那个区间
{
i++;
}
digit=a-sum[i-1];
temp=0;
for(i=1;temp<digit;i++)
{
temp+=(int)log10((double)i)+1;
}
return (i-1)/(int)pow(10.0,temp-digit)%10;//不太理解
}
#include<cstdio>
#include<math.h>
using namespace std;
unsigned int num[31270],sum[31270];
void sequence();
int judge(int a);
int main()
{
int t,n;
sequence();
cin>>t;
while(t--)
{
cin>>n;
cout<<judge(n)<<endl;
}
}
void sequence()
{
int i,j,k;
num[1]=1,sum[1]=1;
for(i=2;i<31270;i++)
{
num[i]=num[i-1]+(int)log10((double)i)+1;//到10时，就会有两位，所以log10保证了位数的准确
sum[i]=sum[i-1]+num[i];//当前总位数
}
}
int judge(int a)
{
int i=1,digit,temp;
while(sum[i]<a)//找在那个区间
{
i++;
}
digit=a-sum[i-1];
temp=0;
for(i=1;temp<digit;i++)
{
temp+=(int)log10((double)i)+1;
}
return (i-1)/(int)pow(10.0,temp-digit)%10;//不太理解
}