UVA - 11621 Small Factors（简单二分）

Computation of Discrete Fourier Transform (DFT) is necessary in many computer programs that
require some kind of digital signal processing, like audio/image processing and spectral analysis. The
most popular algorithm for computing DFT is the Fast Fourier Transform algorithm (FFT), which
takes profit from the factorization into small prime factors of the number of samples that the digital
signal consists of (say, n). The most widely used variant of the FFT algorithm is the so called Radix-2
FFT, which assumes that n is a natural power of 2. Its main drawback is that the number of samples
has to be increased (usually filled with zeroes) to the smallest natural power of 2 that is greater than
or equal to n. This implies that given a number of samples n, the actual number of samples to be
transformed may grow up to 2n, which in practical terms means a 100% computation overload.
An alternative that substantially reduces this upper bound is using a mixed Radix-2/3 FFT algorithm. In this case, the number of samples to be transformed should be expressed as a product of prime
factors 2 and 3:
C2,3 = {n = 2i
· 3
j
, such that i, j belong to N}
Given a positive integer number m, find the smallest number n in the set C2,3, as defined above,
such that n ≥ m. We will denote this number as n = Next2,3(m).
Input
The input consists of a sequence of positive integer numbers, m, one number per line. The end of the
input is marked by a value m = 0. No input number m shall be greater than 231
.
Output
For each non-zero input value m, the program is to write one line with the value of Next2,3(m), as
defined above. No trailing/leading blank spaces should be written after/before any output number.
Sample Input
100
108
1000
3000
0
Sample Output
108
108
1024
3072

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大致题意

在$(2^i\ast 3^j),i,j<=31$中找到大于等于$m$的最小值

林老师发的题，简单二分，用lower_bound解决即可

#include 
using namespace std;
typedef pair<int, int> PII;
#define int long long
const int N = 1e6 + 10;
int a[N];
int phi(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res *= a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
signed main()
{
    int m;
    int cnt = 0;
    for (int i = 0; i < 31; i++)
        for (int j = 0; j < 31; j++)
        {
            int t = phi(2, i) * phi(3, j);
            a[cnt++] = t;
        }
    sort(a, a + cnt);
    while (cin >> m, m)
    {
        int p = lower_bound(a, a + cnt, m) - a;
        cout << a[p] << "\n";
    }
    return 0;
}