【Codeforces】 B. Make it Divisible by 25

B. Make it Divisible by 25

The question ：

Here's a bunch of numbers , Can remove some numbers from it . Ask at least how many numbers to remove , So that the number represented by this series of numbers can be 25 to be divisible by .

Ideas

Can be 25 Divisible number , It should be based on 00、25、50、75 Ending number . The form appearing in the number string should be as follows ： It should be recorded ：（ From right to left ）
… 0 … 0 … ： the second 0, first 0
… 5 … 0 … ： first 0, 0 The first after 5
… 2 … 5 … ： first 5, 5 The first after 2
… 7 … 5 … ： first 5, 5 The first after 7

AC Code ：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int t;
ll n;
int a[20];
int digit(ll i){
    
    int cnt = 0;
    for(int i = 0; i <= 20; i++) a[i] = 0;
    while(i){
    
        a[++cnt] = i % 10;
        i = i / 10;
    }
    reverse(a+1, a+1+cnt);
    return cnt;
}
int main(){
    
    cin >> t;
    while(t--){
    
        cin >> n;
        int num = digit(n);
        int first_0, second_0, first_5, second_5, second_7, second_2;
        first_0 = second_0 = first_5 = second_5 = second_7 = second_2 = 0;
        for(int i = num; i >= 1; i --) {
    
            //00 50 75 25
            //...0...0..
            //...5...0..
            //...7...5..
            //...2...5
            if(a[i] == 0 && (!first_0 || !second_0)){
    
                if(!first_0) first_0 = i;
                else second_0 = i;
            }
            else if(a[i] == 5){
    
                if(!first_5) first_5 = i;
                if(first_0 && !second_5) second_5 = i;
            }
            else if(a[i] == 7){
    
                if(first_5 && !second_7) second_7 = i;
            }
            else if(a[i] == 2){
    
                if(first_5 && !second_2) second_2 = i;
            }
        }
        int ans1, ans2, ans3, ans4, ans;
        ans1 = ans2 = ans3 = ans4 = 0x3f3f3f3f;
        if(first_0 && second_0) ans1 = num - first_0 + first_0 - second_0 - 1;
        if(first_0 && second_5) ans2 = num - first_0 + first_0 - second_5 - 1;
        if(first_5 && second_7) ans3 = num - first_5 + first_5 - second_7 - 1;
        if(first_5 && second_2) ans4 = num - first_5 + first_5 - second_2 - 1;
        ans = min(ans1, min(ans2, min(ans3, ans4)));
        cout << ans << endl;
    }
}