POJ-2661Factstone Benchmark
2024-10-15 23:58:38
Factstone Benchmark
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 5577 | Accepted: 2524 |
Description
Amtel has announced that it will release a 128-bit computer chip by 2010, a 256-bit computer by 2020, and so on, continuing its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in 2000, a 32-bit computer in 1990, a 16-bit computer in 1980, an 8-bit computer in 1970, and a 4-bit computer, its first, in 1960.)
Amtel will use a new benchmark - the Factstone - to advertise the
vastly improved capacity of its new chips. The Factstone rating is
defined to be the largest integer n such that n! can be represented as
an unsigned integer in a computer word.
Given a year 1960 <= y <= 2160, what will be the Factstone rating of Amtel's most recently released chip?
Amtel will use a new benchmark - the Factstone - to advertise the
vastly improved capacity of its new chips. The Factstone rating is
defined to be the largest integer n such that n! can be represented as
an unsigned integer in a computer word.
Given a year 1960 <= y <= 2160, what will be the Factstone rating of Amtel's most recently released chip?
Input
There
are several test cases. For each test case, there is one line of input
containing y. A line containing 0 follows the last test case.
are several test cases. For each test case, there is one line of input
containing y. A line containing 0 follows the last test case.
Output
For each test case, output a line giving the Factstone rating.
Sample Input
1960
1981
0
Sample Output
3
8
Source
OJ-ID:
poj-2661
poj-2661
author:
Caution_X
date of submission:
20191010
tags:
math
description modelling:
给定一个数x(x可以达到256位),找到一个数n满足n!<=x&&(n+1)!>x
major steps to solve it:
1.因为x可以达到256位,所以我们同时对n!<=x和(n+1)!>x去对数
2.得到log(n)+log(n-1)+.......+log(1)<=log(x)&&log(n+1)+log(n)+......+log(1)>log(x)
AC Code:
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n;
double w;
while(~scanf("%d",&n)&&n) {
w=log();
for(int i=;i<=n;i+=) {
w*=;
}
int i=;
double f=;
while(f<w) {
f+=log((double)++i);
}
printf("%d\n",i-);
}
}
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