CF 988E Divisibility by 25 思维 第十二

Divisibility by 25

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer $\mathrm{nn from 11 to 10181018 without\; leading\; zeroes.}$

In one move you can swap any two adjacent digits in the given number in such a way that the resulting number will not contain leading zeroes. In other words, after each move the number you have cannot contain any leading zeroes.

What is the minimum number of moves you have to make to obtain a number that is divisible by $\mathrm{2525?\; Print -1 if\; it\; is\; impossible\; to\; obtain\; a\; number\; that\; is\; divisible\; by 2525.}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 10181\le n\le 1018).\; It\; is\; guaranteed\; that\; the\; first\; (left)\; digit\; of\; the\; number nn is\; not\; a\; zero.}$

Output

If it is impossible to obtain a number that is divisible by $\mathrm{2525,\; print -1.\; Otherwise\; print\; the\; minimum\; number\; of\; moves\; required\; to\; obtain\; such\; number.}$

Note that you can swap only adjacent digits in the given number.

Examples

input

Copy

5071

output

Copy

4

input

Copy

705

output

Copy

1

input

Copy

1241367

output

Copy

-1

Note

In the first example one of the possible sequences of moves is 5071 $\to \to 5701 \to \to 7501 \to \to 7510 \to \to 7150.$

 

 题意： 可以移动每位数的位置，使得移动后的数字是25的倍数，问最小需要移动多少位？

emmmmm，好多特殊情况，wa了很多发

能整除25，则最后两位只能是00,25,50,75

枚举0，2，5，7的个数

然后判断这四种情况，需要注意的是如果处于后面的数在前面了，如果此时另一个数不在末位则移动次数需加一

还有如果从第二位开始有连续的0的话且要移动的数位置在第一位此时移动可能导致开头为0，需要多移动0的个数次，还有特判要用到的0在第二位此时需要加的次数少一

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 3*1e3 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    string s;
    while( cin >> s ) {
        ll a = -1, b = -1, c = -1, d = -1, num = 0, ta;
        for( ll i = 0; i < s.length(); i ++ ) {
            if( s[i] == '0' ) {
                num ++;
                if( num >= 2 ) {
                    ta = a;
                }
                a = i;
            } else if( s[i] == '2' ) {
                b = i;
            } else if( s[i] == '5' ) {
                c = i;
            } else if( s[i] == '7' ) {
                d = i;
            }
        }
        ll ans1 = 1e12, ans2 = 1e12, ans3 = 1e12, ans4 = 1e12;
        bool flag = false;
        //debug(a), debug(b), debug(c),debug(d),debug(num);
        if( a != -1 && c != -1 ) {
            if( a < c ) {
                if( c == s.length() - 1 ) {
                    ans1 = min( s.length() - 1 - a, ans1 );
                } else {
                    ans1 = min( s.length() - 1 - a + s.length() - 2 - c + 1, ans1 );
                }
            } else {
                ans1 = min( s.length() - 1 - a + s.length() - 2 - c, ans1 );
            }
            ll t = 0, i = 1;
            while( s[i] == '0' ) {
                t ++;
                i ++;
            }
            if( s[1] == '0' && ( a == 0 || c == 0 ) ) {
                if( s.length() == 3 ) {
                    flag = true;
                } else {
                    ans1 += t;
                    if( a == 1 ) {
                        ans1 --;
                    }
                }
            }
        }
        if( c != -1 && b != -1 ) {
            if( c < b ) {
                if( b == s.length() - 1 ) {
                    ans2 = min( s.length() - 1 - c, ans2 );
                } else {
                    ans2 = min( s.length() - 1 - c + s.length() - 2 - b + 1, ans2 );
                }
            } else {
                ans2 = min( s.length() - 1 - c + s.length() - 2 - b, ans2 );
            }
            ll t = 0, i = 1;
            while( s[i] == '0' ) {
                t ++;
                i ++;
            }
            if( s[1] == '0' && ( b == 0 || c == 0 ) ) {
                if( s.length() == 3 ) {
                    flag = true;
                } else {
                    ans2 += t;
                }
            }
        }
        if( c != -1 && d != -1 ) {
            if( c < d ) {
                if( d == s.length() - 1 ) {
                    ans3 = min( s.length() - 1 - c, ans3 );
                } else {
                    ans3 = min( s.length() - 1 - c + s.length() - 2 - d + 1, ans3 );
                    //debug(ans);
                }
            } else {
                ans3 = min( s.length() - 1 - c + s.length() - 2 - d, ans3 );
            }
            ll t = 0, i = 1;
            while( s[i] == '0' ) {
                t ++;
                i ++;
            }
            if( s[1] == '0' && ( d == 0 || c == 0 ) ) {
                if( s.length() == 3 ) {
                    flag = true;
                } else {
                    ans3 += t;
                }
            }
        }
        //debug(ans);
        if( num >= 2 ) {
            ans4 = min( ans4, s.length() - 1 - a + s.length() - 2 - ta );
        }
        ll ans = min( min( ans1, ans2 ), min( ans3, ans4 ) );
        if( ans == 1e12 && !flag ) {
            cout << -1 << endl;
        } else {
            cout << ans << endl;
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/l609929321/p/9221787.html