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
5
0
113
110
1000000000
1000000000
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~~
但是我看到没人用 字符串处理+模拟 来写233~~（于是我闲的没事写了些）~~
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； ```cpp #include<bits/stdc++.h> using namespace std; int main(){ char s[10000]; char ss[10000]; cin>>s; int len=strlen(s); for(int i=len-1,j=0;i>=0;i--,j++){ ss[j]=s[i]; } if(ss[0]>=‘5‘) { ss[0]=‘0‘; if(len==1) { ss[len+1]=‘1‘; cout<<ss[len+1]; } for(int i=1;i<=len-1;i++){ if (ss[i]==‘9‘) { ss[i]=‘0‘; continue; } if(ss[i]!=‘9‘) { ss[i]+=1; break; } } for(int i=len-1;i>=0;i--){ cout<<ss[i]; } return 0; } if(ss[0]<=‘4‘) { ss[0]=‘0‘; for(int i=len-1;i>=0;i--){ cout<<ss[i]; } return 0; } return 0; } ```
