Total Amount zoj 2476

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

 

这个题目其实还是挺简单的，但是我就是死活都过不了，哎，还很水啊，

如果拿到省赛上这又是水题一道了，哎，这个题目其实还是很简单的，大数累加

再用一个string的字符串求出的数字字符串和逗号和小数点，然后逆序输出就是了，

就这么简单了。还有就是一些大数累加的知识，凭此程序以后可以搞定以后的大数

累加问题了，hehe

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int temp[100];
int ans[100];
char ten[100];
int i,j,n;

while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
memset(ans,0,sizeof(ans));
int k=0;
for(i=0; i<n; i++)
{
scanf("%s",ten);
memset(temp,0,sizeof(temp));
int l=strlen(ten);
k=0;
for(j=0; j<l; j++)
{
if((ten[l-j-1]<='9')&&(ten[l-j-1]>='0'))
{
//printf("%cten",ten[j]);
temp[k]=ten[l-j-1]-'0';
//printf("%d temp",temp[k]);
k++;
}

}
//for(i=0;i<k;i++)
//printf("%dfdfd ",temp[i]);
for(j=0; j<100; j++)
{
ans[j] = temp[j] + ans[j];
ans[j+1] += ans[j] / 10;
ans[j] = ans[j] % 10;
}
}
for(j=0; j<100; j++)
if(ans[100-1-j]!=0)
break;
int la=100-1-j;
//printf("%d%d\n",la,k);
la=la>=k-1 ? la : k-1;//这个适用于0.00+0.00的那些情况，在那些情况下要用k的值，而非la了
// for(i=0; i<10; i++)
// printf("%d",ans[j]);

string a="";
// printf("%dla\n",la);
for(j=0,i=1; j<=la; j++,i++)
{
a+= ans[j] + 48;
//cout<<a<<endl;
if(j==1)
{
a+= ".";
i=0;
}
if(i%3==0&&i!=0&&i!=la-1)
{
a+= ",";
}

}
a+= "$";
reverse(a.begin(), a.end());//字符串逆转
cout<<a<<endl;

}
return 0;
}