先找到是在哪个集合内,再找到是集合内的哪个元素,最后找到元素的第几位数
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
long long table[100010];
long long psum[100010];
int n=100000;
void maketable()
{
int i,flag=1,x=0;
for(i=1;i<=n;i++)
{
if(i%flag==0)
{
flag*=10;
x++;
}
table[i]=table[i-1]+x;
psum[i]+=table[i]+psum[i-1];
}
}
int main()
{
//freopen("in","r",stdin);
//freopen("out","w",stdout);
maketable();
int T,i,t;
long long x;
char s[100];
cin>>T;
while(T--)
{
cin>>x;
i=lower_bound(psum,psum+n,x)-psum;
if(psum[i]==x)
cout<<i%10<<endl;
else
{
x-=psum[i-1];
i=lower_bound(table,table+n,x)-table;
if(table[i]==x)
cout<<i%10<<endl;
else
{
x-=table[i-1];
sprintf(s,"%d",i);
cout<<s[x-1]<<endl;
}
}
}
return 0;
}
UVA - 10706
option=com_onlinejudge&Itemid=8&page=show_problem&problem=1647" target="_blank">Number Sequence
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second
A single positive integer iis given. Write a program to find the digit located in the position
iin the sequence of number groups S1S2…Sk. Each group
Skconsists 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 <=25), 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 Output for Sample Input
|
2 8 3 |
2 2 |
Problem source: Iranian Contest
Special Thanks: Shahriar Manzoor, EPS.
Source
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::
option=com_onlinejudge&Itemid=8&category=316">
Divide and Conquer -Binary Search
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Divide and Conquer ::
Binary Search