【PAT甲级】1059 Prime Factors (25 分) 题意: 输入一个正整数N(范围为long int),输出它等于哪些质数的乘积。 trick: 如果N为1,直接输出1即可,数据点3存在这样的数据。 如果N本身是一个质数,直接输出它等于自己即可,数据点4存在这样的数据。 AAAAAccepted code: 1#defineHAVE_STRUCT_TIMESPEC2#include<bits/...
1059 Prime Factors (25分) 1. 题目 2. 思路 先求解出int范围内的所有素数,把输入x分别对素数表中素数取余,判断是否为0,如果为0继续除该素数知道余数不是0,遍历到sqrt(x)就够了 3. 注意点 输入数为1的情况, 输出1=1 4. 代码 #include<cstdio> #include<cstring> #include<cmath> #include<...
Factor N in the format N = p1^k1* p2^k2*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent kiis the number of pi-- hence when there is only one pi, kiis 1 and must NOT be printed out. Sample Input: 97532468 Sample Output: 97532468=2^2*11*1...
FactorNin the formatN=p1^k1*p2^k2*…*pm^km, wherepi's are prime factors ofNin increasing order, and the exponentki is the number ofpi -- hence when there is only onepi,ki is 1 and mustNOTbe printed out. Sample Input: 97532468 Sample Output...
primelist[index++] = i; } } } void getprimefactors() { num = 0; for (int i = 0; i < index; ++i) { if (objNum % primelist[i] == 0) { facs[num].x = primelist[i]; facs[num].cnt = 0; while (objNum % facs[num].x == 0) { ...
思路很巧妙,先将所有的数置1,然后在一次i循环中(和上面一样,开方循环),再对所有i*j<50000的循环数i*j置零,因为这些数除1和自身外还有额外的两个因数(不一定是质数)。 #include<vector>vector<int>prime(50000,1);for(inti=2;i*i<50000;i++)for(intj=2;j*i<50000;j++)prime[j*i]=0;...
PAT_甲级_1059 Prime Factors (25分) (C++)【分解因数/规格化输出】,1,题目描述2,思路3,解题过程第一搏万恶的测试点3,#include<iostream>#include<algo
B1059 Prime Factors (25分) 判断是否是素数 素数表打表 质因子分解并存储 #include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<math.h>#include<vector>#include<queue>#include#include<set>#include<stack>#definelowbit(i)((i)...
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Factor N in the format N=p1^k1*p2^k2*…*pm^km , where pi 's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one...