【ZOJ2476】Total Amount（字符串模拟大数相加）

题目链接

Total Amount

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

解题思路：

其实题目本身不难但是要注意细节。

1.数据范围很大一定要用字符串，因为字符串加减不方便，所以这里把字符转存到整数数组里相加

2.还要特判一下两个数相加后整数部分没有0的情况。

3.加，的位置和PAT甲级第一题很像，稍微找一下规律。

代码：

#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;
}
/*
2
$2.45
$3.56
*/