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POJ 1019 Number Sequence

2014-07-15 11:36 274 查看
Number Sequence

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 33873 Accepted: 9691
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

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest


题目大意：

给出一个正整数n，输出在这组序列中第n的位置。

序列：11212312341234512345612345671234567812345678912345678910123456789101112345678910........

解法：利用公式(int)log10((double)i)+1;求出一个数的位数，把每一组分成一个区间，例如1 、12、123、1234、12345、123456、1234567等。第一次循环确定在哪一个大区间，然后再区间中具体找位置。

#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int m;
cin>>m;
int m1;
for(m1=1;m1<=m;m1++)
{
int n;
cin>>n;
__int64 i,j,k;
__int64 len=0,sum=0;
__int64  p;
for(i=1;;i++)
{
p=(int)log10((double)i)+1;
len=len+p;
sum=sum+len;
if(sum>=n)
break;
}
__int64 g=0;
g=n+len-sum;//求在区间中的第几位
__int64 f;
k=0;
while(g>0)
{
k++;
f=(int)log10((double)k)+1;
g=g-f;
}
g=g+f;//多减了一次
char a[1000];
j=0;
while(k>0)
{
a[j]=k%10+48;
k=k/10;
j++;
}
printf("%c\n",a[f-g]);//倒着存入的
}
return 0;
}

                                            

                                        

                        
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