POJ1019 Number Sequence-白红宇

POJ1019 Number Sequence

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发布时间：2019-06-08

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Number Sequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36256		Accepted: 10461

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

Sample Output

Source

, First Iran Nationwide Internet Programming Contest

题意就不说了，看下就明白。思路就是打表，然后就是要知道求一个数的位数：

(int)log10((double)x)+1

/*ID: LinKArftcPROG: 1019.cppLANG: C++*/#include #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   using namespace std;#define eps 1e-8#define randin srand((unsigned int)time(NULL))#define input freopen("input.txt","r",stdin)#define debug(s) cout << "s = " << s << endl;#define outstars cout << "*************" << endl;const double PI = acos(-1.0);const int inf = 0x3f3f3f3f;const int INF = 0x7fffffff;typedef long long ll;const int maxn = 40000;ll sum[maxn], line[maxn];//分别是前i行的位数和，第i行的位数int getbit(int x) { return (int)log10((double)x) + 1;}void init() { sum[1] = line[1] = 1; for (int i = 2; i <= 35000; i ++) { line[i] = line[i-1] + getbit(i); sum[i] = sum[i-1] + line[i]; }}int getpos(int x, int pos) { int len = getbit(x); for (int i = 1; i <= len - pos; i ++) x /= 10; return x % 10;}int main() { init(); int T; ll n; scanf("%d", &T); while (T --) { scanf("%lld", &n); int cur = 1; while (sum[cur] < n) cur ++; ll pos = n - sum[cur-1]; while (cur >= 1) { if (pos > line[cur-1]) { pos -= line[cur-1]; printf("%d\n", getpos(cur, pos)); break; } else cur --; } } return 0;}

转载于:https://www.cnblogs.com/LinKArftc/p/4902512.html

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