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The Super Powers UVA - 11752(数学)_数学题目链接_ltrbless的博客-CSDN博客

题目链接: The Super Powers UVA - 11752 题意: 输出在范围 2^64 - 1范围内的,可以被拆成 至少两种数的n次方形式的数,例如:16 = 2^4 = 4 ^ 2 底数可以为2也可以为4. 题解: 所以一个数的指数只要是合数就可以,因为 (n^a)^b == (n^b
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UVa 11752 - The Super Powers (数论)_zhaosdfa-CSDN博客

A The Super Powers We all know the Super Powers of this world and how they manage to get advantages in political warfare or even in other sectors. But
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UVA 11752 The Super Powers_weixin_33797791的博客-CSDN博客

We all know the Super Powers of this world and how they manage to get advantages in political warfare or even in other sectors. But this is not a pol
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UVA-11752 The Super Powers_weixin_30675247的博客-CSDN博客

We all know the Super Powers of this world and how they manage to get advantages in political warfare or even in other sectors. But this is not a poli
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UVA11752 The Super Powers 素数筛选_Albafica的博客-CSDN博客

求出所有小于2^64-1的Super Powers 一个数n是Super Powers当 a^i=b^j=n 很显然,super powers的幂指数一定是和数,素数筛选找出小于64的所有和数,然后枚举底数在2到1
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The Super Powers UVA - 11752(合数幂)-布布扣-bubuko.com

题意: 求1~2^64-1之间所有的 至少是两个不同的正整数的幂的数 升序输出 一个数的合数次幂即为这样的数 找出1~2^64-1中所有数的合数次幂 用set存起来(既能防止重复 又能升序) 最后输出就好了 #include #include#include
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uva 11752 - The Super Powers(数论+枚举技巧) - 12 - ITeye博客

题目链接:uva 11752 - The Super Powers 题目大意:没有输入,找出所有的超级数,超级数即可以拆分成至少两个数的幂形式。 解题思路:先用素数筛选法找出64以内的合数,以为只有幂是合数才可以进行拆分。然后枚举底数进行判断,所有小于2^64-1的数都是满足的,这里有一个技巧,就是
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uva 11752 The Super Powers 素数+大数判断大小_weixin_30838873的博客-CSDN博客

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2852 题意:找到在[1,2^64-1]区间范围内的所有Super Powers数,Super
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Uva 11752 The Super Powers 解题报告(数学)_SF-_-的博客-CSDN博客

The Super Powers We all know the Super Powers of this world and how they manage to get advantages in political warfare or even in other sectors. But t
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UVA - 11752 The Super Powers_那些年-CSDN博客

We all know the Super Powers ofthis world and how they manage to get advantages in political warfare or evenin other sectors. But this is not a politi
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