codeforces 628B（数学）

题意描述

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.

判断被4整除的子串数量

思路

通过观察发现，百位以上数字的后两位如果可以被4整除，那么这个数字就可以被4整除，所以我们从后往前遍历一遍即可。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=3*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
char s[N];
void solve(){
    scanf("%s",s+1);
    int n=strlen(s+1);
    ll ans=0;
    rep(i,1,n+1){
        int u=s[i]-'0';
        if(u%4==0) ans++;
    }
    rrep(i,n,2){
        int u1=s[i]-'0',u2=s[i-1]-'0';
        if((u2*10+u1)%4==0) ans+=i-1;
    }
    cout<<ans<<endl;
}
int main(){
    //IOS;
    solve();
    return 0;