Luke Skywalker gave Chewbacca an integer number x. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digitt means replacing it with digit 9?-?t.

Help Chewbacca to transform the initial number x to the minimum possiblepositive number by inverting some (possibly, zero) digits. The decimal rePResentation of the final number shouldn't start with a zero.

The first line contains a single integer x(1?≤?x?≤?10¹⁸) — the number that Luke Skywalker gave to Chewbacca.

Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.

27Output
22Input
4545Output
4444
題目大意：
給你一個(gè)LL范圍內(nèi)的數(shù)字X，每一位子上的數(shù)字t都可以替換為9-t，讓你輸出最小的，不含有前導(dǎo)0的正整數(shù)，不能以0開頭。
思路：
貪心角度很好想，ans【i】=min（a【i】，9-a【i】）；
一開始讀題沒看到正整數(shù)三個(gè)字，那么990的ans我認(rèn)為是0.Wa一發(fā)、
后來發(fā)現(xiàn)了，那么990的ans我認(rèn)為是9.又Wa一發(fā)、
哦，結(jié)果要求沒有0開頭啊。那么ans==900.
這是一道A題.恩.滿滿的Hack點(diǎn)，我不禁開始幻想當(dāng)時(shí)打這場(chǎng)比賽的小伙伴們會(huì)Hack多少發(fā)。
Ac代碼：
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define ll __int64char a[200];int ans[200];ll output;int n;int main(){    while(~scanf("%s",a))    {        n=strlen(a);        output=0;        int f=0;        for(int i=0;i<n;i++)        {            ans[i]=min(a[i]-'0',9-a[i]+'0');            if(ans[i]!=0&&f==0)f=1;            if(ans[i]==0&&f==0)ans[i]=max(a[i]-'0',9-a[i]+'0'),f=1;        }        for(int i=0;i<n;i++)        {            output=output*10+ans[i];        }        printf("%I64d/n",output);    }}