UVA-701 The Archeologists' Dilemma （数论）

题目大意：给了一个2^E的前缀n，已知前缀n的位数不到2^E的位数的一半，找出满足条件的最小E。

题目解析：设2^E为i位数，则有n*10^i<2^E<(n+1)*10^i。解不等式得到i*log10(n)/log10(2)<E<i*log10(n+1)/log10(2)。从log10(n)+2开始枚举 i 即可。

代码如下：

# include<iostream>
# include<cstdio>
# include<cmath>
# include<cstring>
# include<algorithm>
using namespace std;
void work(unsigned n)
{
    double u=(double)n;
    int i=log10(u)+2;
    while(1){
        int low=floor(log2(u)+i*log2(10));
        int high=ceil(log2(u+1)+i*log2(10));
        if(high>low+1){
            printf("%d\n",low+1);
            return ;
        }
        ++i;
    }
}
int main()
{
    unsigned n;
    while(scanf("%u",&n)!=EOF)
    {
        work(n);
    }
    return 0;
}

UVA-701 The Archeologists' Dilemma （数论）