CF898A Rounding

题意翻译

给你一个数字，将其“四舍六入”，末尾为5舍去或进位都可，求最终的数字。

题目描述

Vasya has a non-negative integer n n n . He wants to round it to nearest integer, which ends up with 0 0 0 . If n n n already ends up with 0 0 0 , Vasya considers it already rounded.

For example, if n=4722 n=4722 n=4722 answer is 4720 4720 4720 . If n=5 n=5 n=5 Vasya can round it to 0 0 0 or to 10 10 10 . Both ways are correct.

For given n n n find out to which integer will Vasya round it.

输入输出格式

输入格式：

The first line contains single integer n n n ( 0<=n<=109 0<=n<=10^{9} 0<=n<=109 ) — number that Vasya has.

输出格式：

Print result of rounding n n n . Pay attention that in some cases answer isn't unique. In that case print any correct answer.

输入输出样例

输入样例#1：

5

输出样例#1：

0

输入样例#2：

113

输出样例#2：

110

输入样例#3：

1000000000

输出样例#3：

1000000000

输入样例#4：

5432359
输出样例#4： 

5432360

说明

In the first example n=5 n=5 n=5 . Nearest integers, that ends up with zero are 0 0 0 and 10 10 10 . Any of these answers is correct, so you can print 0 0 0 or 10 10 10 .

虽然只是一道四舍五入的水题

我没有看数据范围...于是字符串处理+模拟 来写233

思路（c++）
    1.首先读入字符串    
   2.将字符串倒序存入另一个数组    
   3.此时最后一位数即新数组的第一位     
   4.判断是否大于等于五（ps：我计算的四舍五入）    
   5.如果大于五的话...
        a.先判断是否为一位数（防止“9”这种数据）        
        b.将最后一位数字变为'0'；        
        c.再进行循环（判断下一位是否为9？是 则变为0再继续  否 则这位加一然后退出循环）；        
        d.倒序输出；    
   6.如果小于五...就只把末位变为0，然后倒序输出；

 #include<bits/stdc++.h>
 using namespace std;
 int main(){
     char s[];
     char ss[];
     cin>>s;
     int len=strlen(s);
     for(int i=len-,j=;i>=;i--,j++){
         ss[j]=s[i];
     }
         if(ss[]>='') {
             ss[]='';
             if(len==) {
                 ss[len+]='';
                 cout<<ss[len+];
             }
             for(int i=;i<=len-;i++){
                 if (ss[i]=='') {
                     ss[i]='';
                     continue;
                 }
                 if(ss[i]!='') {
                     ss[i]+=;
                     break;
                 }
             }
             for(int i=len-;i>=;i--){
                     cout<<ss[i];
             }
             return ;
         }
         if(ss[]<='') {
             ss[]='';
             for(int i=len-;i>=;i--){
                 cout<<ss[i];
             }
             return ;
         }

     return ;
 }