UVA 10706 Number Sequence

最新推荐文章于 2022-01-20 17:09:02 发布

coding__girl

最新推荐文章于 2022-01-20 17:09:02 发布

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分类专栏： uva 二分查找算法 c++

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本文介绍了一个算法问题，即在一个由连续整数序列组成的特殊字符串中，如何确定任意指定位置上的具体数字。通过预先构建该序列并记录关键位置信息，文章提供了一种高效的查找方法。

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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 ≤ 25), 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 Outpu
2 2

题意：
有这样一串序列11212312341234512345612345671234567812345678912345678910123456789101112345678910…，问第i个位置数的值。
注意：是第i个位置，而不是第i个数

#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;

class NumberSequence{
private:
    string sk;//储存1-k
    vector<unsigned> position;//储存sk最后一位的位置
    void pushdigit(int i);
public:
    void initial();
    char findNumber(int n);
};
void NumberSequence::initial()
{
    sk.push_back('0');//第一位初始化为0
    position.push_back(0);//第一位初始化为0
    for(int k = 1; position.back() < 2147483647; k++)
    {
        pushdigit(k);
        position.push_back(position.back() + sk.length() - 1);//得到当前总位数
    }
}
void NumberSequence::pushdigit(int k)
{
    stack<int> digit;
    int m=k;
    while(m>0)
    {
        digit.push(m%10);//存入栈中
        m/=10;
    }
    while(!digit.empty())
    {
        sk.push_back(char(digit.top()+'0'));//逆序取出，添加到sk中
        digit.pop();
    }

}
char NumberSequence::findNumber(int n)
{
    vector<unsigned>::iterator it;
    it=lower_bound(position.begin(),position.end(),n);//找到position中第一个比n小的数
    it--;//第一位是0
    return sk[n-(*it)];//返回sk中相应的字符
}
int main(){
    NumberSequence ns;
    ns.initial();
    int c,n;
    char m;
    cin >> c;
    while(c>0)
    {
        cin >> n;
        m=ns.findNumber(n);
        cout << m << endl;
        c--;
    }
    return 0;
}