B. New Skateboard

题目连接：

http://www.codeforces.com/contest/628/problem/A

Description

Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.

You are given a string s consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.

A substring of a string is a nonempty sequence of consecutive characters.

For example if string s is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The only line contains string s (1 ≤ |s| ≤ 3·105). The string s contains only digits from 0 to 9.

Output

Print integer a — the number of substrings of the string s that are divisible by 4.

Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Sample Input

124

Sample Output

4

Hint

题意

给你一个只含有数字的串，然后问你有多少个子串是4的倍数，可以有前导0

题解：

4的倍数的话，当且仅当这个数的个位和十位能被4整除

那么我们扫一遍统计一下就好了，如果这个个位和十位能被4整除，那么他的贡献就是他是整个字符串的第几位

因为他和他的前缀组合起来肯定也是4的倍数

代码

#include<bits/stdc++.h>

using namespace std;

string s;

int main()

{

    cin>>s;

    long long ans = 0;

    for(int i=0;i<s.size();i++)

        if((s[i]-'0')%4==0)

            ans++;

    for(int i=1;i<s.size();i++)

        if(((s[i-1]-'0')*10+(s[i]-'0'))%4==0)

            ans+=1ll*i;

    cout<<ans<<endl;

}