poj 1019 Number Sequence两次二分

Number Sequence Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22614 Accepted: 6031

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 #include<iostream> #include<cstdio> #include<string> #include<algorithm> using namespace std; const int maxn=31270+1; const __int64 inf=2147483647; __int64 a[maxn],cnt,sum[maxn],d[maxn]; __int64 digit(__int64 n) { __int64 cnt=0; while(n) cnt++,n/=10; return cnt; } // 1 2 3 4 5 6 7 8 9 int binarySearch(__int64 a[],int n,__int64 key)//第一个>=key的数的位置 { int l=1,r=n; while(l<r) { int mid=(l+r)/2; if(a[mid]>=key) r=mid; else l=mid+1; } return l; } int main() { for(int i=1;i<maxn;i++) { int dit=digit(i); a[i]=a[i-1]+dit; sum[i]=sum[i-1]+a[i]; } int ci;scanf("%d",&ci); while(ci--) { __int64 n;scanf("%I64d",&n); int p; p=binarySearch(sum,maxn,n); int pos=n-sum[p-1]; int l=binarySearch(a,maxn,pos); int k=pos-a[l-1]; string str; while(l) str+=l+'0',l/=10; reverse(str.begin(),str.end()); cout<<str[k-1]<<endl; } return 0; }