uva 10916 Factstone Benchmark

Jan 29, 2016 · 1 min read · log math

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid;=8&page;=show_problem&problem;=1857 题意：计算最大的n,满足n!/* *********************************************** Author :111qqz Created Time :2016年01月29日星期五 19时49分25秒 File Name :code/uva/10916.cpp ************************************************ */

#include #include #include #include #include #include #include #include #include #include #include #include #define fst first #define sec second #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define ms(a,x) memset(a,x,sizeof(a)) typedef long long LL; #define pi pair < int ,int > #define MP make_pair

using namespace std; const double eps = 1E-8; const int dx4[4]={1,0,0,-1}; const int dy4[4]={0,-1,1,0}; const int inf = 0x3f3f3f3f; const int N=150; int n; int f[N]; void pre() { f[0] = 3; int per = 2; for ( int i = 1 ; i <=25 ; i++) { f[i]=f[i-1]+per; per++; } } int main() { #ifndef ONLINE_JUDGE freopen("code/in.txt","r",stdin); #endif

while (scanf("%d",&n)!=EOF)  // (1*2*3*...*n)>2^k,巧妙利用对数求解。两边取对数即可。
{
    if (n==0) break;
    int k = (n-1960)/10;
    int p = 1<<(k+2);
   // cout<<"k:"<<k<<endl;
   // cout<<"p:"<<p<<endl;
    double dp = p*1.0*log(2); //因为不等式左边每一个都有一个log(2)(计算对数的时候用换底公式产生的)，可以乘到不等号右边。
    double sum = 0 ;
    
    for ( int i = 1; ; i++)
    {
	sum +=log(i);
	//cout<<"i:"<<i<<" sum:"<<sum<<endl;
	if (sum>dp)
	{
	    printf("%d\n",i-1);
	    break;
	}
    }

}

#ifndef ONLINE_JUDGE
fclose(stdin); #endif return 0; }