UVa 1586 - Molar mass

1586 - Molar mass Time limit: 3.000 seconds

For example, the molar mass of a molecular formula C6H5OH is 94.108 g/mol which is computed by6 × (12.01 g/mol) + 6 × (1.008 g/mol) + 1 × (16.00 g/mol).

Given a molecular formula, write a program to compute the molar mass of the formula.InputYour program is to read from standard input.

The input consists of T test cases. The number of testcases T is given in the first line of the input. Each test case is given in a single line, which containsa molecular formula as a string. The chemical symbol is given by a capital letter and the length ofthe string is greater than 0 and less than 80. The quantity number n which is represented after thechemical symbol would be omitted when the number is 1 (2 ≤ n ≤ 99).

Output

Your program is to write to standard output. Print exactly one line for each test case. The line shouldcontain the molar mass of the given molecular formula.

Sample Input

4

C

C6H5OH

NH2CH2COOH

C12H22O11

Sample Output

12.010

94.108

75.070

342.296

解题：

• 定义4个数组，s放要输入的字符串，c放元素的原子质量（打表），遍历数组s，a放元素，b放数字；

• 注意每次循环后要把a、b清零，不然会有错误！！

• 并且注意只有数组b是int型，其他的都是浮点型（sum等），注意占位符格式！！

#include<stdio.h>

#include<string.h>

#include<ctype.h>


int main()

{

    int t;

    scanf("%d",&t);

    int b[85];

    char a[85],s[85];

//    memset(s,0,sizeof(s));


    double c[100];

    c['C']=12.01;c['H']=1.008;c['O']=16.00;c['N']=14.01;

    while(t--)

    {    

        memset(a,0,sizeof(a));

        memset(b,0,sizeof(b));

        double sum=0;

        scanf("%s",s);//printf("%s\n",s);

        for(int i=0;i<strlen(s);i++)

        {

//            printf("s[%d]=%c\n",i,s[i]);

            if(s[i]>='0'&&s[i]<='9')

                b[i]=s[i]-'0';

            else

                a[i]=s[i];

        }

        for(int i=0;s[i];i++)

        {

            if(b[i+1]!=0)

            {

//                printf("b[%d]=%d\n",i+1,b[i+1]);

                if(b[i+2]!=0)

                    {

//                        printf("%lf\n",c[a[i]]);

//                        printf("b[]=%d,%d\n",b[i+1],b[i+2]);

//                        printf("num=%lf\n",b[i+1]*10.0+b[i+2]);

//                        printf("%lf\n",c[a[i]]);

                        sum+=c[a[i]]*(b[i+1]*10.0+b[i+2]);

                        i++;    

                    }

                else {

//                    printf("%lf,%d\n",c[a[i]],b[i+1]);

                    

                    sum+=c[a[i]]*b[i+1];

                }

            }

            else sum+=c[a[i]];

        }

        printf("%.3lf\n",sum);

    }

    return 0;

}