POJ-2661Factstone Benchmark
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.
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
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:
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(4); for(int i=1960;i<=n;i+=10) { w*=2; } int i=1; double f=0; while(f<w) { f+=log((double)++i); } printf("%d\n",i-1); } }