NEC Programming Contest 2021(AtCoder Beginner Contest 229) B - Hard Calculation

题目链接：B - Hard Calculation (atcoder.jp)

You are given positive integers A and B.
Let us calculate A+B (in decimal). If it does not involve a carry, print Easy; if it does, print Hard.

• A and B are integers.
• 1≤A,B≤1018

Input is given from Standard Input in the following format:

A B

If the calculation does not involve a carry, print Easy; if it does, print Hard.

229 390

Hard

When calculating 229+390, we have a carry from the tens digit to the hundreds digit, so the answer is Hard.

123456789 9876543210

Easy

We do not have a carry here; the answer is Easy.
Note that the input may not fit into a 32-bit integer.

题意：问a+b的过程中是否产生进位，如果有输出hard，否则easy

思路：每一位都提取出来来进行运算。直到a为0或者b为0

#include
using namespace std;


int main(){
	long long a, b;
	cin >> a >> b;
	bool flag = 0;
	while(a && b && !flag){
		int ans = a % 10;
		int cnt = b % 10;
		if(ans + cnt >= 10){
			flag = 1;
			break;
		}
		a /= 10;
		b /= 10;
	}
	if(flag){
		cout << "Hard" << endl;
	}else{
		cout << "Easy" << endl;
	}
	return 0;
}