Codeforces Round 1004 (Div. 2) C. Devyatkino

C. Devyatkino

题目链接👈

题目描述🥰

题目思路😀

这个题目其实是不难的，但是正解的解法感觉很新奇🤔

首先我们可以观察到我们的答案肯定是小于等于9的，所以我们在0到9次操作进行枚举，如果某次操作时某位已经到达7了，那么就可以直接返回操作数了。

Question：怎么确定是否有可能将由k个九组成的数字精确相加，产生一个数字7？

要简单回答像 999999..99这样的加法会如何影响一个任意数的数位是相当困难的，但是这个解法的精妙之处就是在于把这个9999...999当做成10^x-1，这样实际上我们进行k次操作就是在n-k的基础上加上k个10的幂

这个问题解决了，那么题目就得到了大大简化，我们直接枚举操作数k，计算n-k的数位里小于7中最接近7的数字res，如果这个res+k>=7,就相当于可以达到7，可以直接退出枚举了。

 

AC代码🧠

// Problem: C. Devyatkino
// Contest: Codeforces - Codeforces Round 1004 (Div. 2)
// URL: https://codeforces.com/contest/2067/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define dev1(a) cout << #a << '=' << a << endl;
#define dev2(a, b) cout << #a << " = " << a << "  " << #b << " = " << b << endl;
#define dev3(a, b, c) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << endl;
#define dev4(a, b, c, d) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << endl;
#define dev5(a, b, c, d, e) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << "  " << #e << " = " << e << endl;
#define vec(a)                         \
    for (int i = 0; i < a.size(); i++) \
        cout << a[i] << ' ';           \
    cout << endl;
#define darr(a, _i, _n)               \
    cout << #a << ':';                \
    for (int ij = _i; ij <= _n; ij++) \
        cout << a[ij] << ' ';         \
    cout << endl;           
#define cin(a,n)           \
     for(int i=0;i<n;i++) \
      cin>>a[i];          
#define endl "\n"
#define pow pim
int pim(int a,int k)
{
    int res=1;
    if(a==0)return 0;
    while(k) 
    {
        if(k&1)res=(int)res*a;
        k>>=1;
        a=(int)a*a;
    }
    return res;
}
#define fi first
#define se second
#define caseT \
    int T;    \
    cin >> T; \
    while (T--)
// #define int long long
// #define int __int128

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 99999999;

// const int N = ;
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
 
int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
void solve()
{
	int n;
    cin >> n;
    //最多有9次操作
    for (int i = 0; i<= 9; i++) {
        string s = to_string(n - i);
        int ans = 0;
        for (char c: s) {
            if (c <= '7') {
                ans = max(ans, c - '0');
            }
        }
        if (i +ans >= 7) {
            cout << i <<endl;
            return;
        }
    }
}

signed main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	caseT
    solve();
    return 0;
}
/*

*/