1059. Prime Factors (25)

2019-11-08 19:24:57

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特殊处理2，然后一次对x/3，/5。。。判断，知道x变成0

#include<iostream>#include<cmath>using namespace std;int main(){ long int x; cin >> x; cout << x << "="; if (x == 0 || x==1) {//额外处理0,1,的情况 cout << x << endl;exit(0); } int count = 0;int flag = 0; while (x % 2 == 0) { //先把2处理掉 count++; x = x / 2; flag = 1; } if (count >1) cout << "2^" << count; else if (count == 1) cout << "2"; for (int i = 3;x!=1;i += 2)//按3.5.7...顺序循环，知道x变成1 { count = 0; while (x%i == 0) { count++; x /= i; } if (flag == 1) { if (count > 1) PRintf("*%d^%d", i, count); else if (count == 1)printf("*%d", i); } else { if (count > 1) printf("%d^%d", i, count); else if (count == 1)printf("%d", i); flag = 1; } } cout << endl;}

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