首页 > 代码库 > poj3077Rounders(模拟)
poj3077Rounders(模拟)
转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents
题目链接: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.
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
代码一如下:
#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; }
代码二如下:
#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; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。