POJ1019:Number Sequence

http://poj.org/problem?id=1019

题目:

Number Sequence

Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 24168Accepted: 6466

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
 
数学类问题,题意是给一串 1 12 123 1234 12345 123456 。。。。这样的数字问第

i个数字是多少。这题的trick在于第i个数与第i-1个数的位数差不定,可以利用公式:(int)log10(double(i))+1,当增加的数字是10的幂次关系时个数会变化。

打表:一开始没有打表,超时了~~~~呵呵····终于知道什么叫打表了!!

#include<math.h>
#include
<iostream>
using namespace std;
unsigned
int a[31270],s[31270];
void reset()//打表
{
int i;
a[
1]=1;
s[
1]=1;
for(i=2;i<31270;i++)
{
a[i]
=a[i-1]+(int)log10((double)i)+1;
s[i]
=s[i-1]+a[i];
}
}

int main()
{
int T;
int n;
int i;
scanf(
"%d",&T);
reset();
while(T--)
{
scanf(
"%d",&n);
i
=1;

while(s[i]<n) i++;

int pos=n-s[i-1];
int tmp=0;
for(i=1;tmp<pos;i++)
{
tmp
+=(int)log10((double)i)+1;
}
int k=tmp-pos;
printf(
"%d\n",(i-1)/(int)pow(10.0,k)%10) ;/*从右向左求,比如123456,k=2,则结果为4*/

}
return 0;

}

这道题是比较简单的了~~~~~~

    原文作者:kuangbin
    原文地址: https://www.cnblogs.com/kuangbin/archive/2011/07/21/2113279.html
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