GDUFE ACM-1356

题目：http://acm.gdufe.edu.cn/Problem/read/id/1356

Problem B. Relative atomic mass

Time Limit: 2000/1000ms (Java/Others)

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 1/12 of 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

思路：判断遇到的字符是哪一个，如果遇到'H'则+1，如果遇到'C'则+12，如果遇到'O'则+16

难度：非常简单

代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int n,i,sum,k;
    char a[11];
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            sum=0;
            scanf("%s",&a);
            k=strlen(a);
            for(i=0;i<k;i++)
            {
                if(a[i]=='H')
                    sum+=1;
                else if(a[i]=='O')
                    sum+=16;
                else if(a[i]=='C')
                    sum+=12;
            }
            printf("%d
",sum);
        }
    }
    return 0;
}