Number Sequence - UVa 10706_uva number sequence-优快云博客

本文提供了一个程序，用于在特定数字序列中找到指定位置的数字。通过输入测试案例的数量和位置索引，程序能够输出对应位置的数字。

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Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second

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…S_k. Each group S_k consists of a sequence of positive integer numbers ranging from 1 to k, written one after another. For example, the first80 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 Output for Sample Input

题意：按照给定的序列，求出第i个数字是什么。

思路：sum[i]表示前i轮所达到的长度，sum[i]=sum[i-1]+len(i)，然后每次，只要让i减去前面的轮数的长度即可，这里也可以用二分，更快一些。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int T,t,num[210000],length;
ll sum[210000],n;
int len(int k)
{
    int ret=0;
    while(k>0)
    {
        ret++;
        k/=10;
    }
    return ret;
}
int main()
{
    int i,j,k,ret;
    num[1]=1;
    sum[1]=1;
    i=1;length=1;
    while(sum[i]<=2147483647LL)
    {
        i++;
        k=len(i);
        ret=i;
        for(j=length+k;j>length;j--)
        {
            num[j]=ret%10;
            ret/=10;
        }
        length+=k;
        sum[i]=sum[i-1]+length;
    }
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%I64d",&n);
        k=1;
        while(n>sum[k])
            k++;
        n-=sum[k-1];
        printf("%d\n",num[n]);
    }
}