poj1019 大数据处理 分块
发布时间:2020-12-14 03:36:38 所属栏目:大数据 来源:网络整理
导读:Number Sequence Time Limit: ?1000MS ? Memory Limit: ?10000K Total 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
Number Sequence
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 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; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |