点击打开链接
Description:
There is a positive integer X, X's reversion count is Y. For example, X=123, Y=321; X=1234, Y=4321. Z=(X-Y)/9, Judge if Z is made up of only one number(0,1,2...9), like Z=11,Z=111,Z=222,don't consider '+'and '-'.
Input:
Input contains of several test cases. Each test case only contains of a number X, L is the length of X. ( 2 <= L < 100)
Output:
Output “YES”or “NO”.
样例输入
10 13
样例输出
YES YES
题目来源
2018 ACM-ICPC 中国大学生程序设计竞赛线上赛
||||||：开始用C++写的一直过不了，后来队友提醒X的长度可达100，用long long（不到20位）也存不开，于是用java翻译了一遍，AC#include<iostream>
#include<queue>
#include<string.h>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>using namespace std;bool judge(int a)
{int t=a%10;while(a){if(t!=a%10)return false;a/=10;}return true;
}int main()
{std::ios::sync_with_stdio(false);cin.tie(0);int x,t,z;while(cin>>x){t=x;int y=0;while(x){y=y*10+x%10;x/=10;}z=(t-y)/9;if(z<0)z=-z;//cout<<x<<"**"<<y<<"z="<<z<<endl;if(judge(z))cout<<"YES"<<endl;elsecout<<"NO"<<endl;}return 0;
}import java.math.BigInteger;
import java.util.Scanner;public class Main {public static void main(String args[]){Scanner sc=new Scanner(System.in);BigInteger x,t,y,z;while(sc.hasNext()){x=sc.nextBigInteger();t=x;y=BigInteger.ZERO;while(!t.equals(BigInteger.ZERO)){y=y.multiply(BigInteger.valueOf(10)).add(t.mod(BigInteger.valueOf(10)));t=t.divide(BigInteger.valueOf(10));}z=(x.subtract(y)).divide(BigInteger.valueOf(9));t=z.mod(BigInteger.valueOf(10)) ;int flag=1;while(!z.equals(BigInteger.ZERO)){if(!(t.equals(z.mod(BigInteger.valueOf(10))))){flag=0;break;}z=z.divide(BigInteger.valueOf(10));}if(flag==1)System.out.println("YES");elseSystem.out.println("NO");}}
}