POJ 3077-Rounders

Rounders

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7970 Accepted: 5156

Description

For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to the nearest hundred, then (if that result is greater than 1000) take that number and round it to the nearest thousand, and so on ...

Input

Input to this problem will begin with a line containing a single integer n indicating the number of integers to round. The next n lines each contain a single integer x (0 <= x <= 99999999).

Output

For each integer in the input, display the rounded integer on its own line.

Note: Round up on fives.

Sample Input

91514459912345678444444451445446

Sample Output

20104510010000000500000002000500题意：输入数字把他变成里他最近的整10或者整100或整1000类推，例如15变成20,14变成10,99变成100,999变成1000注意：虽然是个水题但是首位是9的情况折腾了我好久，感觉我一定智障了
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int main(){    int kase;    cin>>kase;    while(kase--)    {        string x;        cin>>x;        for(int i=x.length()-1;i>0;i--)  //第一位不变        {            if( '5'<=x[i] && x[i]<='9' )  //进位            {                x[i-1]++;            }            x[i]='0';  //两种情况都要变成0        }        if(x[0]>'9')  //当99的情况如果不处理会输出:0        {            cout<<1;            x[0]='0';        }        cout<<x<<endl;    }    return 0;}