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  • Number Sequence

    2020-03-22 21:43:17
    Number Sequence Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 65 Accepted Submission(s) : 27 Problem Description A number sequence is defined...

    Number Sequence
    Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 65 Accepted Submission(s) : 27
    Problem Description
    A number sequence is defined as follows:

    f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

    Given A, B, and n, you are to calculate the value of f(n).

    Input
    The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

    Output
    For each test case, print the value of f(n) on a single line.

    Sample Input
    1 1 3
    1 2 10
    0 0 0

    Sample Output
    2
    5

    Author
    CHEN, Shunbao

    Source
    ZJCPC2004

    #include <stdio.h>
    int fn(int a,int b,int n);
    int main()
    {
        int n,a,b;
        scanf("%d %d %d",&a,&b,&n);
        while(a!=0||b!=0||n!=0)
    	{
          a=a%7;
          b=b%7;
          n=n%48;
          printf("%d\n",fn(a,b,n));
          scanf("%d %d %d",&a,&b,&n);
        }
        return 0;
    }
    int fn(int a,int b,int n)
    {
        if(n==1) 
          return 1;
        else if(n==2) 
          return 1;
        else
          return (a*fn(a,b,(n-1))+b*fn(a,b,(n-2)))%7;
    }
    
    展开全文
  • number sequence

    2018-04-10 19:42:17
    Number SequenceTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 41681 Accepted: 12158DescriptionA single positive integer i is given. Write a program to find the digit located in the ...
    Number Sequence
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 41681 Accepted: 12158

    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

    log10(n)+1可以得到数n的长度

    Len[n]表示1到n这一段的总长度

    s[n]表示第一段到第n段的总长度

    先确定第n位在哪段中,然后从1开始枚举确定第n位在哪个数中。设k是这个数最后一位的位置,这个数为num,则第n位的数是num/pow(10,k-n)%10 (掐头去尾)

    #include <iostream>  
    #include <cstdio>  
    #include <cstring>  
    #include <algorithm>  
    using namespace std;  
    #define LL long long  
    LL s[50005];  
    LL Len[50005];  
      
    #include <cmath>  
    int main(){  
        Len[1]=1;  
        for(int i=2;i<=50000;i++){  
            Len[i]=Len[i-1]+log10(i*1.0)+1;  
        }  
        s[1]=1;  
        for(int i=2;i<=50000;i++){  
            s[i]=s[i-1]+Len[i];  
        }  
        LL n;  
        int T;  
        scanf("%d",&T);  
        while(T--){  
            scanf("%lld",&n);  
            int i;  
            for(i=1;s[i]<n;i++);  
            i--;  
            n-=s[i];  
            int pos;  
            int k=0;  
            for(pos=1;n>k;pos++){  
                k+=(log10(pos*1.0)+1);  
            }  
            pos--;  
            int res;  
            res=pos/(int)pow(10*1.0,(k-n)*1.0);  
            res%=10;  
            printf("%d\n",res);  
              
        }  
        return 0;  
    }  

    展开全文
  • NumberSequence

    2008-04-02 17:40:32
    how to use ax Number Sequence

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