zoj 2476 Total Amount

Given a list of monetary amounts in a standard format, please calculate the total amount.

We define the format as follows:

1. The amount starts with ‘$’.

2. The amount could have a leading ‘0’ if and only if it is less then 1.

3. The amount ends with a decimal point and exactly 2 following digits.

4. The digits to the left of the decimal point are separated into groups of three by commas (a group of one or two digits may appear on the left).

Input

The input consists of multiple tests. The first line of each test contains an integer N (1 <= N <= 10000) which indicates the number of amounts. The next N lines contain N amounts. All amounts and the total amount are between 0.00and 20,000,000.00, inclusive. N=0 denotes the end of input.

Output

For each input test, output the total amount.

Sample Input

2
1,234,567.89 9,876,543.21
3
0.01 0.10
$1.00
0

Sample Output

$11,111,111.10

1.11分析：字符串处理时，分别用两个数组保留小数点前的数，和小数点后面的数，相加，数组每个位置只保留一位数（便于打，）。还要注意打“，”时要注意不要出现类似“ ,123”开始打点情况。
顺便提醒自己写代码时要适当给注意事项注释下，本人开始注意到了不开始打“，”，后面修改代码有给忘了，然后又找了好久bug,无语了！

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
long long a[50],b[50];
char s[50],s2[50];
int main()
{
    int t;
    while(scanf("%d",&t),t)
    {
        int len,ca=0,cb=0,flag2=0;
        char *c;
        memset(a,0,sizeof(a)),memset(b,0,sizeof(b));

        for(int i=0;i<t;i++)
        {
            int fa=0,k=0;
            scanf("%s",s);
            len=strlen(s);
            int m=len;
            if((c=strchr(s,'.'))!=NULL)//小数点出现 
            {
                    m=c-s;
                    flag2=1;
                    for(int j=m+1;j<len;j++)//b数组保留小数点后面的数， 
                    {
                        b[k++]+=s[j]-'0';
                    }
            }

            for(int j=m-1;j>=0;j--) //a数组保留小数点前面的数 
            {
                if(isdigit(s[j]))
                a[fa++]+=s[j]-'0';
            }
            ca=max(ca,fa-1);   //保留小数点前数的最长位置 
        }

        if(b[1]>=10) //小数点后面的数进位 
        b[0]+=b[1]/10, b[1]%=10;

        if(b[0]>=10)
        a[0]+=b[0]/10, b[0]%=10;
        for(int i=0;i<=ca;i++) //小数点前面的数进位 只保留一位数 
        {
            if(a[i]>=10)
            {
                a[i+1]+=a[i]/10;
                a[i]%=10;
            }
        }
        while(a[ca+1]>0)
        {
            ca++; 
            a[ca+1]+=a[ca]/10;
            a[ca]%=10; 
        }
        int flag=1,sum=0,cnt=0;
        printf("$");
            for(int i=0;i<=ca;i++) //分点 
            {
                ++sum;  
                s2[cnt++]=a[i]+'0';
                if(i==ca)
                continue;
                if(sum==3)
                {
                    s2[cnt++]=',';
                    sum=0;
                }
            }
            s2[cnt]=0;
            for(int i=cnt-1;i>=0;i--)
            printf("%c",s2[i]);
        printf(".");
        for(int i=0;i<2;i++)
        printf("%d",b[i]);
        printf("\n");
    }
    return 0;
 }