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;
}

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;  
}