PTA B1048 数字加密 (20 分)

题目链接：点击这里

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>

using namespace std;
typedef long long ll;
const int MOD = 10000007;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 1010;
int a[maxn];

int main()
{
	string a,b,ans;
	cin>>a>>b;
	reverse(a.begin(), a.end());
	reverse(b.begin(), b.end());
	int lena = a.length();
	int lenb = b.length();
	int len = lena>lenb?lena:lenb;	//a和b的较大长度 
	for(int i=0;i<len;i++)
	{
		int numa = i<lena?a[i]-'0':0;	//numa对应a[i]，长度不足时用0代替 
		int numb = i<lenb?b[i]-'0':0;	//numb对应b[i]，长度不足时用0代替
		if(i%2==0)	//字符串的偶数位，即原数的奇数位 
		{
			int temp = (numa+numb)%13;
			if(temp==10)	ans += 'J';
			else if(temp==11)	ans += 'Q';
			else if(temp==12)	ans += 'K';
			else	ans +=  temp+'0';
		}
		else	//字符串的奇数位，即原数的偶数位 
		{
			int temp = numb - numa;
			if(temp<0)	ans+=temp+10+'0';
			else	ans+=temp+'0';
		}
	}
	reverse(ans.begin(), ans.end());
	cout<<ans<<endl;
	return 0;
}

标签：lenb,lena,20,int,PTA,long,const,include,B1048
来源： https://blog.csdn.net/qq_42815188/article/details/100921989