HDU 5347(MZL's chemistry-打表)

MZL's chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 629    Accepted Submission(s): 449

Problem Description

MZL define F(X) as the first ionization energy of the chemical element X

Now he get two chemical elements U,V,given as their atomic number,he wants to compare F(U) and F(V)

It is guaranteed that atomic numbers belongs to the given set:{1,2,3,4,..18,35,36,53,54,85,86}

It is guaranteed the two atomic numbers is either in the same period or in the same group

It is guaranteed that x≠y

 

Input

There are several test cases

For each test case,there are two numbers u,v,means the atomic numbers of the two element

 

Output

For each test case,if F(u)>F(v),print "FIRST BIGGER",else print"SECOND BIGGER"

 

Sample Input

1 2
5 3

 

Sample Output

SECOND BIGGER
FIRST BIGGER

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 5

 

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打表

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int a[24]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,35,36,53,54,85,86} ; 
double f[24]={1312.0,2372.3,520.2,899.5,800.6,1086.5,1402.3,1313.9,1681,2080.7,495.8,737.7,577.5,786.5,1011.8,999.6,1251.2,1520.6,1139.9,1350.8,1008.4,1170.4,890,1037};
double h[1000];
int main()
{
//	freopen("E.in","r",stdin);
	
	Rep(i,24) h[a[i]]=f[i];
	
	int u,v;
	while (cin>>u>>v)
	{
		if (h[u]<h[v]) puts("SECOND BIGGER");
		else puts("FIRST BIGGER");
	}
	
	
	return 0;
}