UVa 10392 - Factoring Large Numbers
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传送门UVa 10392 - Factoring Large Numbers
因式分解的题目.
实在想不到什么好方法, 只好用简单粗暴的方法过.
注意题目中的这句话
You may assume that there will be at most one factor more than 1000000.
也就是说除了数本身以外所有的因子都小于1000000.
这样循环就有一个限度了.
详情见代码
#include <cstdio>#include <cmath>#include <cstring>using namespace std;int main(){ //freopen("input.txt", "r", stdin); register long long i, n, k; while (scanf("%lld", &n)) { if (n < 0) break; //我把输入的数分两种情况处理, 仅仅是为了提高一点点的时间效率. 去掉任何一种都对. if ((int)sqrt(n) < 1000000) { for (i = 2; i <= (int)sqrt(n); i++) { while (n % i == 0) { printf(" %lld\n", i); n /= i; } if (n == 1) break; } if (n > 1) printf(" %lld\n", n); } else { for (i = 2; i <= 1000000; i++) { while (n % i == 0) { printf(" %lld\n", i); n /= i; } if (n == 1) break; } } if (i > 1000000) printf(" %lld\n", n); printf("\n"); } return 0;}
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