【21.67%】【codeforces 727B】Bill Total Value

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Vasily exited from a store and now he wants to recheck the total price of all purchases in his bill. The bill is a string in which the names of the purchases and their prices are printed in a row without any spaces. Check has the format “name1price1name2price2…namenpricen”, where namei (name of the i-th purchase) is a non-empty string of length not more than 10, consisting of lowercase English letters, and pricei (the price of the i-th purchase) is a non-empty string, consisting of digits and dots (decimal points). It is possible that purchases with equal names have different prices.

The price of each purchase is written in the following format. If the price is an integer number of dollars then cents are not written.

Otherwise, after the number of dollars a dot (decimal point) is written followed by cents in a two-digit format (if number of cents is between 1 and 9 inclusively, there is a leading zero).

Also, every three digits (from less significant to the most) in dollars are separated by dot (decimal point). No extra leading zeroes are allowed. The price always starts with a digit and ends with a digit.

For example:

“234”, “1.544”, “149.431.10”, “0.99” and “123.05” are valid prices,

“.333”, “3.33.11”, “12.00”, “.33”, “0.1234” and “1.2” are not valid.

Write a program that will find the total price of all purchases in the given bill.

Input

The only line of the input contains a non-empty string s with length not greater than 1000 — the content of the bill.

It is guaranteed that the bill meets the format described above. It is guaranteed that each price in the bill is not less than one cent and not greater than 106 dollars.

Output

Print the total price exactly in the same format as prices given in the input.

Examples

input

chipsy48.32televizor12.390

output

12.438.32

input

a1b2c3.38

output

6.38

input

aa0.01t0.03

output

0.04

【题解】



一开始的时候一直在纳闷样例1是啥。

看了挺久还剩一个半小时，就先去做T3了，A了T3之后再回过头来看(本来是想说题目错了会改，但是并没有等到);

然后才看到。如果dolar那一位有三个以上的数字需要用点号分割.

12.123指的是12123美元

而

12.12指的是12美元12美分

然后就累加和呗。找到每个数字串里面最后一个点号的位置(如果存在);

然后看一下那个点号后面有几个数字，如果有3个，则整个字串全是dolar位，否则有两位小数(表示美分),注意最多只有两位小数！；

最后写到输出答案的部分的时候比赛时间只剩5分钟。

我想说肯定够了。

突然发现自己不会输出double x = 0.32的小数部分；

怎么把点号去掉呢？

我想着乘10，然后取整，再减去整数，再乘10再取整！(但是竟然爆精度了,玄学);

最后比赛完的那一瞬间猛然想到直接乘100再去整数部分不就好了！

等终测完了，代码把乘100的代码交上去，A了。哎万念俱灰。如果····，可惜没有如果。

当然不能直接取整数部分，如果小于10要再补个0；

哎，TAT掉了24分。明天还要上高数辅导班，得早退了。不然不能打DIV2了。我梦寐以求的重归蓝名计划(没错我蓝过！！)离我越来越远。

都怪我一直在捣鼓T3的交互是啥玩意0 0MLGB,现在想想要是T3知道交互输入是什么就能节省很多时间了。说不定这题就A了，啊啊啊啊啊啊啊啊啊！！

#include <string>
#include <iostream>
#include <algorithm>

using namespace std;

string s;
double ans = 0;

int main()
{
//  freopen("F:\\rush.txt", "r", stdin);
    cin >> s;
    int len = s.size();
    int now = 0;
    while (now <= len - 1)//这整个while用于获取字符串里面包含的数字的和
    {
        int j = now + 1;
        if (isdigit(s[now]))
        {
            double num = 0;
            while (j <= len-1 && isdigit(s[j]) || s[j] == '.')
                j++;
            int t = j-now;
            string s1 = s.substr(now, t);
            int len1 = s1.size();
            int pd = -1;
            for (int i = len1-1;i >= 0;i--)
                if (s1[i] == '.')
                {
                    pd = i;//pd是这个数字len1里面最后一个点号的位置
                    break;
                }
            int size = len1-1 - pd;//这个点号后面的数字个数
            if (pd==-1 || size > 2)//全是dolars
            {
                for (int i = 0; i <= len1 - 1; i++)
                    if (isdigit(s1[i]))
                        num = num * 10 + (s1[i] - '0');
                ans += num;//累加答案
            }
            else//有美分在后面的情况
            {
                for (int i = 0; i <= pd - 1; i++)
                    if (isdigit(s1[i]))
                        num = num * 10 + (s1[i] - '0');
                double tt = 1;
                for (int i = pd + 1; i <= len1 - 1; i++)
                {//小数部分
                    tt = tt*0.1;
                    num += (s1[i] - '0')*tt;
                }
                ans += num;//累加答案
            }
        }
        now = j;
    }
    double temp = int(ans);
    int dd = int(ans);
    int no = 0;
    string ss = "";//输出的时候美元也要3位一点号，有点麻烦所以用字符串吧
    while (dd > 0)
    {
        int key = dd % 10;
        char ke = key + '0';
        ss = ke + ss;
        no++;
        dd = dd / 10;
        if (no == 3)
        {
            no = 0;
            if (dd >0)
                ss = '.' + ss;
        }
    }
    if (abs(ans - temp) <= 0.001)//如果没有小数部分
    {
        cout << ss << endl;
    }
    else//有小数部分
    {
        if (ss == "")
            printf("0");
        else
            cout << ss;
        printf(".");
        double tt = ans - temp;
        tt = tt * 100;//乘100···，说多了都是泪
        if (tt < 10)
            printf("0");
        printf("%.0lf\n", tt);
    }
    return 0;
}