ZR993

首先,这种和平方有关的,首先应当考虑根号做法

这道题目,我们可以直接暴力\(\log_{10}w + 10\)判断一个数是否能够由原数变化的到

直接\(O(\sqrt{n})\)枚举所有的平方数,直接暴力上面的方法check

#include<cstdio>

#include<iostream>

#include<queue>

#include<algorithm>

#include<cstring>

#include<cctype>

#include<vector>

#include<ctime>

#include<map>

#define LL long long

#define pii pair<int,int>

#define mk make_pair

#define fi first

#define se second

using namespace std;

const int N = 55;

const LL INF = 1e13;

LL n;

int s[N];

int num[11];

int num2[11];

int h[N];

inline LL read(){

	LL v = 0,c = 1;char ch = getchar();

	while(!isdigit(ch)){

		if(ch == '-') c = -1;

		ch = getchar();

	}

	while(isdigit(ch)){

		v = v * 10 + ch - 48;

		ch = getchar();

	}

	return v * c;

}

inline bool work(LL x){

	memset(num2,0,sizeof(num2));

	if(x == 0) num2[0]++;

	while(x){

		num2[x % 10]++;

		x /= 10;

	}

	for(int i = 0;i < 10;++i) if(num2[i] > num[i]) return 0;

	return 1;

}

int main(){

	n = read();

	while(n){

		memset(num,0,sizeof(num));

		LL g = n;int t = 0;

		while(g){

			h[++t] = g % 10;

			num[g % 10]++;

			g /= 10;

		}

		sort(h + 1,h + t + 1);

		LL to = 1;

		for(int i = t;i >= 1;--i) to = to * 10 + h[i];

		for(LL i = 0;1ll * i * i <= to;++i){

			if(work(i) && work(i * i))

				printf("%lld * %lld = %lld\n",i,i,1ll * i * i);

		}

		n = read();

	}

	return 0;

}

巴特西

ZR993

ZR993

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