4 条题解
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2cxccxc LV 9 @ 2018-12-28 17:26:34
打表出奇迹
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cctype> #include<algorithm> #include<stack> #include<queue> #include<vector> #include<map> #define INF 1<<30 #define LL long long #define ULL unsigned long long using namespace std; int p[]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0}; int n; void f(int x) { for(int i=2;i<=x;i++) { if(p[i]==1&&x%i==0) { cout<<i; if(x/i!=1) { cout<<"*"; f(x/i); } return; } } } int main() { cin>>n; cout<<n<<"="; f(n); return 0; }
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02024-08-15 20:38:57@
#include <iostream> using namespace std; // 递归函数:分解质因数并打印 void factorize(int n, int divisor) { // 如果 n 已经变成1,说明分解完成 if (n == 1) { return; } // 如果 n 能被当前的 divisor 整除 if (n % divisor == 0) { // 打印这个质因数 cout << divisor; // 继续分解剩下的部分 if (n / divisor != 1) { cout << "*"; } factorize(n / divisor, divisor); } else { // 如果不能被当前的 divisor 整除,检查下一个数 factorize(n, divisor + 1); } } int main() { int number; cout << "请输入一个合数: "; cin >> number; cout << number << "="; factorize(number, 2); // 从2开始分解 cout << endl; return 0; }
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02020-12-02 21:51:11@
def dg(t,n,k): if t == True: if n % k == 0 and n / k != 1: print (k,end="*") n = n / k dg(t,n,k) elif n % k != 0 : k += 1 dg(t,n,k) else: print(k,end="") t = False n = int(input()) t = True print(n,end="=") dg(t,n,2)
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02020-08-18 13:54:26@
感觉好像有点问题,不过判出来是对的
import math
def fj(x):
f=False
for i in range(2,int(math.sqrt(x)//1)+1):
if x%i==0:
f=True
break
if f==False:
print(int(x),end='')
else:
print(i,end='*')
fj(x/i)x=int(input())
print(str(x)+'=',end='')
fj(x)
- 1
信息
- 难度
- 6
- 分类
- (无)
- 标签
- (无)
- 递交数
- 218
- 已通过
- 58
- 通过率
- 27%
- 被复制
- 1
- 上传者