poj1019 大数据处理 分块

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33215 Accepted: 9490

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

283

Sample Output

2

粗看想暴力过的，看到20几亿的数据不用想时间肯定超时。昨晚把边界数据大小数了下，本想很繁的题目，今晚真正上机写完估计不用5分钟，当然代码质量可能不是很高，可以更精简些。提交后RE了好几次，就很郁闷了。最后发现一个变量打错，丢死人

记录一下统计结果，当然如果有更好的写法这些数据可能用不上

1~9        共45位

10~99      共9000位               45+9000=9045

100~999    共1386450位            45+9000+1386450=1395495

1000~9999  共188019000位          45+9000+1386450+188019000=189414495

10000~9999 共23750235000位        

第2147483647位于第31268个1...n中

本题用分块方法写的，也许是最笨的写法，有其他更好写法，欢迎指导

#include<iostream>#include<cmath>using namespace std;long a[31270]={0},t,sum=0,i,j,k,len,h,g;long b[490000]={0};long f(long n,long x){len=0;for(g=n;g>=1;g--){h=g;while(h){b[len++]=h%10;h/=10;}}return b[len-x];}int main(){for(i=1;i<10;i++){a[i]=a[i-1]+1;sum+=a[i];}for(i=10;i<100;i++){a[i]=a[i-1]+2;sum+=a[i];}for(i=100;i<1000;i++){a[i]=a[i-1]+3;sum+=a[i];}for(i=1000;i<10000;i++){a[i]=a[i-1]+4;sum+=a[i];}for(i=10000;i<100000&&sum<=2147483647;i++){a[i]=a[i-1]+5; sum+=a[i];}    cin>>t;if(t<1||t>10)return 0;while(t--){cin>>i;if(i<1||i>2147483647)return 0;k=i;if(i>189414495){k-=189414495;  for(j=10000;;j++){  if(k>a[j]){    k-=a[j];continue;  }  else{  cout<<f(j,k)<<endl;  break;  }  }}else if(i>1395495){k-=1395495;for(j=1000;;j++){if(k>a[j]){k-=a[j];continue;}else{cout<<f(j,k)<<endl;      break;}}}else if(i>9045){k-=9045;for(j=100;;j++){if(k>a[j]){k-=a[j];continue;}else{cout<<f(j,k)<<endl;      break;}}}else if(i>45){k-=45;for(j=10;;j++){  if(k>a[j]){k-=a[j];continue;  }  else{cout<<f(j,k)<<endl;      break;  }}}else if(i>=1){for(j=1;;j++){  if(k>a[j]){k-=a[j];continue;  }  else{cout<<f(j,k)<<endl;      break;  }}}}  return 0;}