poj 3077 Rounders
2014-10-14 20:20
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Rounders
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
Sample Output
题目意思:给你一个数字,按照它给的规则求出它的近似数;规则:对于445,先转化为450,再转化为500;对于123456,》123460》123500》12400》12000》10000;
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7656 | Accepted: 4961 |
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
9 15 14 4 5 99 12345678 44444445 1445 446
Sample Output
20 10 4 5 100 10000000 50000000 2000 500
题目意思:给你一个数字,按照它给的规则求出它的近似数;规则:对于445,先转化为450,再转化为500;对于123456,》123460》123500》12400》12000》10000;
#include <iostream> using namespace std; int main(){ int t,n,num[10]; cin>>t; while (t--){ cin>>n; if (n<=10) cout<<n<<endl; else{ int k=0; while (n>0){ num[k++]=n%10; n/=10; } for (int i=0;i<k-1;i++){ if (num[i]>=5) num[i+1]++; num[i]=0; } while (k--) cout<<num[k]; cout<<endl; } } return 0; }
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