Codeforces Round #426 (Div. 2) C The Meaningless Game

题意
检测两个数是否可以表示成a=k1^2*k2+k3*k4^2 b = k1*k2^2*k3^2*k4
即一个是平方,一个就是原本的.
我推了一些简单的必要条件,进行堆叠,勉强过了,但是具体还不知道,留待后考吧.

/* Farewell. */
#include <iostream>
#include <vector>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <set>
using namespace std;
#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#define gcd __gcd
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define MP make_pair
#define MT make_tuple
#define PB push_back
typedef long long
LL;
typedef unsigned long long ULL;
typedef pair<int,int > pii;
typedef pair<LL,LL> pll;
typedef pair<double,double > pdd;
typedef pair<double,int > pdi;
const int INF = 0x7fffffff;
const LL INFF = 0x7f7f7f7fffffffff;
const int MOD = 1e9+7;
#define debug(x) std::cerr << #x << " = " << (x) << std::endl
const int MAXM = 5e3+17;
const int MAXN = 1e6+17;
LL cub[MAXN];
int main(int argc, char const *argv[])
{
#ifdef GoodbyeMonkeyKing
freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
for (int i = 0; i < MAXN; ++i)
{
cub[i] = 1LL*(i+1)*(i+1)*(i+1);
}
int t;
cin>>t;
while(t--)
{
LL a,b;
scanf("%lld%lld",&a,&b);
LL temp = a*b;
//debug(temp);
//debug(cub[lower_bound(cub, cub+MAXN, temp)-cub]);
if(cub[lower_bound(cub, cub+MAXN, temp)-cub]!=temp)
{
puts("NO");
continue;
}
LL oa = a,ob = b;
if(a==b)
{
if(a==1)
{
puts("YES");
continue;
}
if(cub[lower_bound(cub, cub+1002, a)-cub]==a)
puts("YES");
else
puts("NO");
continue;
}
LL gd = gcd(a,b);
a/=gd;
b/=gd;
if(gd%(a*b)==0)
{
puts("YES");
}
else puts("NO");
}
return 0;
}

最后

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