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題目鏈接：http://poj.org/problem?

id=3077

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

9 15 14 4 5 99 12345678 44444445 1445 446

Sample Output

20 10 4 5 100 10000000 50000000 2000 500

代碼一、例如以下：

#include <iostream> #include <cstring> using namespace std; int main() {int t, n, k, count;char s[17];int i, j;while(cin >>t){while(t--){count = 0;int p = 0, l = 0;;memset(s,0,sizeof(s));cin>>s;int len = strlen(s);if(len == 1){cout<<s[0]<<endl;continue;}for(i = len-1; i > 0; i--){if(s[i]-'0'+p > 4){p = 1;count++;}else{p = 0;count++;}}if(s[0]-'0' + p > 9){cout<<10;}else{cout<<s[0]-'0'+p;}for(i = 0; i < count; i++){cout<<'0';}cout<<endl;}}return 0; }
代碼二、例如以下：

#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> using namespace std;int main() {int t;scanf("%d", &t);while (t--){int n, count = 0;scanf("%d", &n);double x = n;while (x >= 10){x /= 10;x = (int)(x + 0.5);count++;}n = (int)x;for (int i = 0; i < count; i++)n *= 10;printf("%d\n", n);}return 0; }

轉載于:https://www.cnblogs.com/hrhguanli/p/5192996.html

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