UVa 10916 factstone Benchmark: Techniques for handling math and factorial

10916-factstone Benchmark

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem &problem=1857

Amtel has announced that it'll release a 128-bit computer chip by, a 256-bit computer by 2020, and, Continuin G its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in), a 32-bit computer in 1990, a 16-bit computer in 1980, a 8-bit computer in 1970 , and a 4-bit computer, its-alpha, in 1960.)

Amtel would use a new benchmark-the factstone -to advertise the vastly improved the capacity of it new chips. The Factstone rating is defined to being largest integern such that n! can be represented as a n unsigned integer in a computer word.

Given a year 1960≤y≤2160, what would be the Factstone rating of Amtel ' s most recently-released?

There are several test cases. For each test case, the there is one line of input containing y. A line containing 0 follows of the last Test case. For each test case, output a line giving the Factstone rating.

Sample Input

1960
1981
0

Output for Sample Input

3
8

Train of thought: The number of bytes k = (year-1940)/10, the problem is turned into n! < 2 ^ K < (n + 1)!, if the simple simulation will overflow, so we take the logarithm on both sides, because log (a*b) = log (a) + log (b), so log (n!) = SUM (log (i)), (1<= i <= n ), just find the smallest sum (log (i)) > K * log (2), the answer is i-1.

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

/*0.286s*/
  
#include <cstdio>
#include <cmath>
const double Log_2=log (2.0);
  
int main (void)
{
    int year;
    while (scanf ("%d", &year), year)
    {
        int n = (year-1940)/ten;
        Double k = POW (2, N) * log_2, sum = 0;
        for (int i = 1;; i++)
        {
            sum = log (i);
            if (Sum > K)
            {
                printf ("%d\n", i-1);
                break;
            }
    }} return 0;
}

Play table:

/*0.012s*/
  
#include <cstdio>
  
const int ANS[21] =
{
    3, 5, 8, A,
    170,
    536, 966, 1754, 3210, 5910, 10944, 20366, 38064, 71421, 134480, 254016
};
  
int main (void)
{
    int year;
    while (scanf ("%d", &year), year)
        printf ("%d\n", ans[(year-1960)/ten);
}