New Skateboard_CodeForces_628B

最新推荐文章于 2021-01-13 23:24:43 发布

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最新推荐文章于 2021-01-13 23:24:43 发布

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分类专栏： ACM 数学

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本文介绍了一个算法问题，即在给定的数字字符串中找出所有能被4整除的子串数量。通过分析，我们发现只需检查每两个连续数字组成的子串是否为4的倍数，同时考虑单个数字的情况。最后，提供了一段C++代码实现这一解决方案。

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题目：

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.

Examples

Input

Output

Input

Output

Input

5810438174

Output

题意：

给你一个字符串，然后让你求是4的倍数的子串（连续的）有几个。

题解：

给你一个串fdgdfdab 每个字母代表不同的数字，假设4可以整除ab，那么4也可以整除dab，fdab，.....以此往前，因为：
拿这个字符串当例子：fdgdfdab % 4 = （fdgdfd * 100 + ab）% 4 == fdgdfd00 % 4 + ab % 4 == 0 + ab % 4 所以当ab是4的倍数时fdgdfdab可以整除以4。所以只需要判断连续的两个数字是不是4的倍数，是的话那么以这两个数字结尾的子串都可以整除以4。不要漏掉一个数字就可以乘除的情况。一共这两种情况。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<sstream>
#include<vector>
#include<map>
#include<set>
#define up(i, x, y) for(int i = x; i <= y; i++)
#define down(i, x, y) for(int i = x; i >= y; i--)
using namespace std;
typedef long long ll;
int n, m, k;
int a, b;
ll T, t;
ll ans = 0;
int main()
{
    string str; cin>>str;
    if((str[0] - '0') % 4 == 0) ans++; //先算了单独第一个
    up(i, 1, str.length() - 1)
    {
        char a, b;
        a = str[i - 1];  
        b = str[i];       // ab为连续的两个数字
        t = (a - '0') * 10 + (b - '0');  // 把字符ab转成整型
        if((b - '0') % 4 == 0) ans++;   //看单个的数字可不可以除尽4
        if(t % 4 == 0){                 //看ab是否可以整除以4
            ans += i;
        }
    }
    cout<<ans<<endl;
}