1059 Prime Factors (25 分)质因子易错题

1059 Prime Factors (25 分)

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

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 pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

具体代码

#include <cstdio>
#include <cmath>bool isprime(int a){if(a<=1)return false;int sqr=(int)sqrt(a);for(int i=2;i<=sqr;i++){if(a%i==0)false;}return true;}int prime[100020],pnum=0;//pnum为100020以内质数的个数void findprime(){for(int i=0;i<100020;i++){if(isprime(i)==true)prime[pnum++]=i;}}//0到n内的所有质数，并传入数组prime[]struct factor{int x,cnt;//x为质因子，cnt为质因子的个数
}fac[15];int main(){findprime();int n,num=0;//num为不同的质因子的个数scanf("%d",&n);if(n==1)printf("1=1");else{printf("%d=",n);int sqr=sqrt(n);//因为n的因子只可能在根号n以内for(int i=0;i<pnum&&prime[i]<sqr;i++){if(n%prime[i]==0){fac[num].x=prime[i];fac[num].cnt=0;while(n%prime[i]==0){fac[num].cnt++;n/=prime[i];}num++;}if(n==1)break;}if(n!=1){fac[num].x=n;fac[num++].cnt=1;}for(int i=0;i<num;i++){if(i>0)printf("*");            printf("%d",fac[i].x);if(fac[i].cnt>1)printf("^%d",fac[i].cnt);}}return 0;}

1059 Prime Factors (25 分)质因子易错题相关推荐

【题目分析】1059 Prime Factors (25 分)
立志用最少的代码做最高效的表达 PAT甲级最优题解-->传送门 Given any positive integer N, you are supposed to find all of its ...
1059 Prime Factors (25 分)【难度: 一般 / 知识点: 分解质因子】
注意: 判断1000000007这种只有一个质数特别大的情况 #include<bits/stdc++.h> using namespace std; map<int,int> ...
1059. Prime Factors (25)
题目如下: Given any positive integer N, you are supposed to find all of its prime factors, and write the ...
PAT甲题题解-1059. Prime Factors (25)-素数筛选法
用素数筛选法即可. 范围long int,其实大小范围和int一样,一开始以为是指long long,想这就麻烦了该怎么弄. 而现在其实就是int的范围,那难度档次就不一样了,瞬间变成水题一枚,因为i ...
1059. Prime Factors (25)--taste
点击打开链接注意: 1.对于输入为1的情况要特判 2.factor在int范围内开到10就足够 3.sqrt是范围,由于n后面被修改故范围要单写,且sqrt是double int sqr=sqrt( ...
用计算机计算一定比笔算简便,四年级数学(下册)期末知识要点汇总，易错题解析...
当我第一遍读一本好书的时候,我仿佛觉得找到了一个朋友;当我再一次读这本书的时候,仿佛又和老朋友重逢.我们要把读书当作一种乐趣,并自觉把读书和学习结合起来,做到博览.精思.熟读,更好地指导自己的学习,让 ...
PAT甲级1059 Prime Factors ：[C++题解]分解质因子
文章目录题目分析题目链接题目分析暴力求质因数下面i就是质因子,s是质因子i的阶数. 暴力的时间复杂度O(n),会超时 void divide(int n){for(int i=2;i< ...
1059 Prime Factors（25 分)(cj)
1059 Prime Factors(25 分) Given any positive integer N, you are supposed to find all of its prime fac ...
1059 Prime Factors
1059 Prime Factors (25 分) Given any positive integer N, you are supposed to find all of its prime fa ...
PAT 1059 Prime Factors[难]
1059 Prime Factors (25 分) Given any positive integer N, you are supposed to find all of its prime fa ...

1059 Prime Factors (25 分)质因子易错题

Input Specification:

Output Specification:

Sample Input:

Sample Output:

1059 Prime Factors (25 分)质因子易错题相关推荐

最新文章

热门文章

1059 Prime Factors (25 分)质因子 易错题

Input Specification:

Output Specification:

Sample Input:

Sample Output:

1059 Prime Factors (25 分)质因子 易错题相关推荐

最新文章

热门文章

1059 Prime Factors (25 分)质因子易错题

1059 Prime Factors (25 分)质因子易错题相关推荐