uva 10061 How many zero's and how many digits ?
2013-09-29 11:22
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这道题稍微有一点麻烦,求总共占多少个位数比较好解决利用对数就可以轻松得出答案,但是要求最后由多少个0,需要分析一下问题,只要在阶乘的过程中出现了一个基数,就会在最后的结果里面多一个0,所以先将基数分解质因数,看每个质因数究竟需要多少个才能相乘得出一个基数,然后对参与阶乘的n个数分别进行分解,能够提取出基数的质因数的,就将质因数提取出来,统计所有可能提取出的质因数的个数,根据这个统计结果对比形成每一个基数需要的质因数的个数,就可以得出至多能够构成的基数的个数,自然就是最后阶乘的结果中最后包含的0的个数了。我觉得这个思路比较笨,不是一种快速的计算方法,小弟初出茅庐,技能拙劣,欢迎牛人指教更好的方法。
#include <stdio.h> #include <math.h> #include <map> using namespace std; int prime_table[801]; typedef int PRIME_FACTOR; typedef int NUMBER; map<PRIME_FACTOR, NUMBER> m, m1; bool is_prime(int n) { int i, end; bool f; if(1 == n) return false; end = floor(sqrt((double)n)+0.5); f = true; for(i=2; i<=end; i++) { if(0 == n%i) { f = false; break; } } return f; } //初始化素数表 void init_prime_table() { int i; int count; count = 0; for(i=1; i<=800; i++) { if(is_prime(i)) prime_table[count++] = i; } } //找出n的所有质因数 void get_factor(int n) { int i; m.clear(); for(i=0; i<=800&&prime_table[i]&&prime_table[i]<=n; i++) { if(n == 1) break; if(0 == n%prime_table[i]) { n = n/prime_table[i]; if(m.find(prime_table[i]) != m.end()) { m[prime_table[i]] ++; i = -1; } else { m[prime_table[i]] = 1; i = -1; } } } } void func(int n, int base) { get_factor(base); m1.clear(); map<PRIME_FACTOR, NUMBER>::iterator it; /* for(it=m.begin(); it!=m.end(); it++) { printf("factor=%-10d number=%-10d\n", it->first, it->second); } */ int i; int t, min; for(i=1; i<=n; i++) { t = i; for(it=m.begin(); it!=m.end(); it++) { loop: if(t == 1) break; if(t%it->first == 0) { t = t/it->first; if(m1.find(it->first) != m1.end()) { m1[it->first] ++; it = m.begin(); goto loop; } else { m1[it->first] = 1; it = m.begin(); goto loop; } } } } min = (m1.begin())->second / m.begin()->second; for(it=m.begin(); it!=m.end(); it++) { t = (int)(m1[it->first]/it->second); if( min > t ) min = t; } //上面的代码是计算最后有几个0,下面的代码利用对数的性质计算结果总共占多少位 double sum; sum = 0; for(i=1; i<=n; i++) sum += log((double)i)/log(base); printf("%d %.0lf\n", min, floor(sum)+1); } int main(void) { int n, base; init_prime_table(); while(scanf("%d %d",&n,&base) != EOF) { func(n, base); } return 0; }
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