Codeforces 628 B.New Skateboard

 
B. New Skateboard
 
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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
124
Output
4
Input
04
Output
3
Input
5810438174
Output
9
 

题意就是找有多少个子串可以被4整除。因为100可以被4整除，所以100位以上的都可以，只要看个位和十位的数是否可以被4整除。
写这个题的时候傻了，只是算一下个位和十位就好了，然后把前面的位数加上就可以了，。
因为想一下，举个例子，xyzab，如果ab%4==0，那么xyzab可以组成多少个子串呢？那就是ab，zab，yzab，xyzab，就是a所在的位置啊，因为是从0开始的，所以位数+1，我可能是个傻子。。。
这样这道题就可以a了。

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=3*1e5+10;
 5 char a[N];
 6 int main(){
 7     while(~scanf("%s",a)){
 8     ll len,cnt,ans;
 9     len=strlen(a);
10     ans=0;
11     for(int i=0;i<len;i++){
12         if((a[i]-'0')%4==0)ans++;
13     }
14     for(int i=0;i<len-1;i++){
15         cnt=(a[i]-'0')*10+(a[i+1]-'0');
16         if(cnt%4==0)ans+=i+1;
17     }
18     cout<<ans<<endl;
19     }
20     return 0;
21 }

 








posted @ 2017-07-27 19:45  ZERO-  阅读(331)  评论(0)  编辑  收藏  举报