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

283

Sample Output

22

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。