HDU 5347 MZL's chemistry (2015 Multi-University Training Contest 5 2015多校联合)

最新推荐文章于 2015-08-23 20:58:46 发布

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本篇博客介绍了一个简单的程序设计问题，即通过给定的化学元素的原子序数来比较两种元素的第一电离能大小。通过预先设定好的第一电离能数值表，程序能够快速准确地给出比较结果。

题目传送门：HDU 5347 MZL's chemistry

MZL's chemistry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 32 Accepted Submission(s): 29

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

Source

2015 Multi-University Training Contest 5

高中生出的题= =。啊。找来那个第一电离能的表直接打表就出结果了。

代码：

#include"iostream"
#include"cstdio"
using namespace std;
int main()
{
    int u,v;
    int a[104]={1312.0,2372.3,520.2,899.5,800.6,1086.5,1402.3,1313.9,1681.0,2080.7,
                695.8,737.7,577.5,786.5,1011.8,999.6,1251.2,1520.6,418.8,589.8,
                633.1,658.8,650.9,652.9,717.3,762.5,760.4,737.1,745.5,906.4,
                57.8,762.0,794.0,941.0,1139.9,1350.8,403.0,549.5,900.0,640.1,
                652.1,684.3,702.0,710.2,719.7,804.4,781.0,867.8,558.3,708.6,
                834.0,869.3,1008.4,1170.4,375.7,502.9,538.1,534.4,527.0,533.1,
                540.0,544.5,547.1,593.4,565.8,573.0,581.0,589.3,596.7,603.4,
                523.5,658.5,761.0,770.0,760.0,840.0,880.0,870.0,890.1,1007.1,
                589.4,715.6,703.0,812.1,890.0,1037.0};
    while(scanf("%d%d",&u,&v)!=EOF)
    {
        if(a[u-1]>a[v-1])
        {
            printf("FIRST BIGGER\n");
        }
        else
        {
            printf("SECOND BIGGER\n");
        }
    }

    return 0;
}