Number Sequence--组合数学

Number Sequence Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 37890 Accepted: 10952

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 题目链接：http://poj.org/problem?id=1019

题目的意思很明确吧，我就不用再说了吧。

组合数学这种东西搞得我很是尴尬啊，用一个数组预处理一下位数，恩，没错，然后找到具体的数，然后，就没有然后了。。

代码

#include <cstdio> #include <iostream> #include <cstring> #include <cmath> using namespace std; long long b[35001]; long long a[35001]; void fax()//预处理位数 { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); for(int i=1;i<=35000;i++) { a[i]=a[i-1]+(int )log10((double)i)+1; b[i]=b[i-1]+a[i]; } } int solve(int x) { int i; for(i=0;i<=32000;i++)//找到恰好的数 { if(x<=b[i+1]&&x>b[i]) break; } x-=b[i]; int t=0,k; for(i=1;;i++) { k=(int)log10((double)i)+1; t+=k; if(t>=x) break; }//计算具体的是哪一位 if(t==x) return i%10; else { int kk=t-x; int h=1; while(kk--) h*=10; return (i/h)%10; } } int main() { int x,t; scanf("%d",&t); fax(); while(t--) { scanf("%d",&x); cout<<solve(x)<<endl; } return 0; }

希望明天的比赛成绩会好吧。