ZOJ 2476 Total Amount 字符串模拟

- Total Amount

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2476

Description

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 $0.00 and $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

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e+5+5;
char a[maxn];
int c[maxn],d[maxn],ans[maxn];
int main()
{
    int T;
    while(~scanf("%d",&T) && T)
    {
        memset(ans,0,sizeof(ans));
        memset(d,0,sizeof(d));
        int index,cnt=0;
        while(T--)
        {
            memset(c,0,sizeof(c));
            memset(a,0,sizeof(a));
            scanf("%s",a);
            int k=0;
            for(int i=0; i<strlen(a); i++)
            {
                if(isdigit(a[strlen(a)-i-1]))
                    c[k++]=a[strlen(a)-i-1]-'0';
            }
            for(int i=0; i<100; i++)
            {
                d[i]+=c[i];
                d[i+1]+=d[i]/10;
                d[i]%=10;
            }
        }
        for(int i=100; i>=0; i--)
        {
            if(d[i]!=0)
            {
                index=i;
                break;
            }
        }
        for(int i=index; i>=2; i--)
            ans[cnt++]=d[i];
        printf("$");
        if(cnt==0)printf("0");
        else
        {
            for(int i=0; i<cnt; i++)
            {
                printf("%d",ans[i]);
                if((cnt-i-1)%3==0 && i!=cnt-1)
                    printf(",");
            }
        }
        printf(".%d%d\n",d[1],d[0]);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/yinbiao/p/9494636.html