题解 | #硬币兑换#

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    
    int coins[] = {2, 5, 10, 20, 50, 100};
    int typeCount = 1;  // 已经包含1元硬币
    int remaining = n - 1;  // 减去一个1元硬币
    int totalCoins = n;  // 初始假设全用1元
    
    // 尝试使用每种面值的硬币
    for(int i = 0; i < 6; i++) {
        if(remaining < coins[i]) {
            break;
        }
        typeCount++;
        remaining -= coins[i];
        totalCoins = totalCoins - coins[i] + 1;
    }
    
    cout << typeCount << " " << totalCoins << endl;
    return 0;
}

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        int[] coins = {2, 5, 10, 20, 50, 100};
        int typeCount = 1;  // 已经包含1元硬币
        int remaining = n - 1;  // 减去一个1元硬币
        int totalCoins = n;  // 初始假设全用1元
        
        // 尝试使用每种面值的硬币
        for(int i = 0; i < 6; i++) {
            if(remaining < coins[i]) {
                break;
            }
            typeCount++;
            remaining -= coins[i];
            totalCoins = totalCoins - coins[i] + 1;
        }
        
        System.out.println(typeCount + " " + totalCoins);
    }
}

n = int(input())

coins = [2, 5, 10, 20, 50, 100]
type_count = 1  # 已经包含1元硬币
remaining = n - 1  # 减去一个1元硬币
total_coins = n  # 初始假设全用1元

# 尝试使用每种面值的硬币
for coin in coins:
    if remaining < coin:
        break
    type_count += 1
    remaining -= coin
    total_coins = total_coins - coin + 1

print(f"{type_count} {total_coins}")