HDU 5949 Relative atomic mass 【模拟】 （2016ACM/ICPC亚洲区沈阳站）

Relative atomic mass

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 66    Accepted Submission(s): 59

Problem Description

Relative atomic mass is a dimensionless physical quantity, the ratio of the average mass of atoms of an element (from a single given sample or source) to 12of the mass of an atom of carbon-12 (known as the unified atomic mass unit).
You need to calculate the relative atomic mass of a molecule, which consists of one or several atoms. In this problem, you only need to process molecules which contain hydrogen atoms, oxygen atoms, and carbon atoms. These three types of atom are written as ’H’，’O’ and ’C’ repectively. For your information, the relative atomic mass of one hydrogen atom is 1, and the relative atomic mass of one oxygen atom is 16 and the relative atomic mass of one carbon atom is 12. A molecule is demonstrated as a string, of which each letter is for an atom. For example, a molecule ’HOH’ contains two hydrogen atoms and one oxygen atom, therefore its relative atomic mass is 18 = 2 * 1 + 16.

 

Input

The first line of input contains one integer N(N ≤ 10), the number of molecules. In the next N lines, the i-th line contains a string, describing the i-th molecule. The length of each string would not exceed 10.

 

Output

For each molecule, output its relative atomic mass.

 

Sample Input

5 H C O HOH CHHHCHHOH

 

Sample Output

1 12 16 18 46

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5949

题目大意：

　　给一个只含C H O的分子式，求相对分子质量。

题目思路：

　　【模拟】

　　水题，直接模拟即可。

//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-8)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#define N 24
#define M 1004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
char s[N];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("1.txt","r",stdin);
//    freopen("2.txt","w",stdout);
    #endif
    int i,j,k;
    int x,y,z;
//    init();
    for(scanf("%d",&cass);cass;cass--)
//    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//    while(~scanf("%s",s))
//    while(~scanf("%d%d",&n,&m))
    {
        scanf("%s",s);
        n=strlen(s);ans=0;
        for(i=0;i<n;i++)
        {
            if(s[i]=='H')ans++;
            else if(s[i]=='O')ans+=16;
            else if(s[i]=='C')ans+=12;
        }
        printf("%d
",ans);
    }
    return 0;
}
/*
//

//
*/