CF813B：The Golden Age（数学）

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本文介绍了一个算法问题，目标是在给定区间内找到最长的连续GoldenAge，即不存在由特定形式表示的不幸年份的时期。通过预先计算并排序可能的不幸年份，使用二分查找等技术高效求解。

Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.

For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.

Such interval of years that there are no unlucky years in it is called The Golden Age.

You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.

Input

The first line contains four integer numbers x, y, l and r (2 ≤ x, y ≤ 1018, 1 ≤ l ≤ r ≤ 1018).

Output

Print the maximum length of The Golden Age within the interval [l, r].

If all years in the interval [l, r] are unlucky then print 0.

   Examples
 
     input
   
2 3 1 10

     output
   
1

     input
   
3 5 10 22

     output
   
8

     input
   
2 3 3 5

     output
   
0

Note

In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].

In the second example the longest Golden Age is the interval [15, 22].

题意：找出区间内最长的连续Golden Age。

思路：INT64最多只有二的六十多次方，所以可以将所有数枚举出来，排一排序即可。

# include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[5000], b[70], c[70];
int main()
{
    int cnt=0;
    LL x, y, l, r, ans = 0;
    scanf("%lld%lld%lld%lld",&x,&y,&l,&r);
    for(LL t=1LL; t>0&&t<=r; t*=x) {b[++b[0]] = t; if(r*1.0/t<x) break;}
    for(LL t=1LL; t>0&&t<=r; t*=y) {c[++c[0]] = t; if(r*1.0/t<y) break;}
    for(int i=1; i<=b[0]; ++i)
        for(int j=1; j<=c[0]; ++j)
            a[cnt++] = b[i]+c[j];
    sort(a, a+cnt);
    cnt = unique(a, a+cnt)-a;
    int p = lower_bound(a, a+cnt, l) - a;
    if(p == cnt || a[p] > r) return 0*printf("%lld\n",r-l+1);
    ans = a[p]-l;
    for(++p; p<cnt&&a[p]<=r; ++p)
        ans = max(ans, a[p]-a[p-1]-1);
    ans = max(ans, r-a[p-1]);
    printf("%lld\n",ans);
    return 0;
}