POJ 2883 Checking order

时间限制: 
1000ms
内存限制: 
65536kB
描述
输入长度为5的数字串,检查是否是按从小到大的顺序排列,如果是,输出Yes;否则,输出No,并输出从小到大排序后的结果。
输入
长度为5的一串数字;
输出
判断结果及排序后的结果。
样例输入
1 3 5 7 9
5 3 44 7 3
1 1 2 2 3
样例输出
Yes
No 3 3 5 7 44
Yes
 
(1)、源代码:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
 
int main(){
                int i;
                vector<int> v(5);
 
                while(cin >> v[0] >> v[1] >> v[2] >> v[3] >> v[4]){
                                if(v[0]<=v[1] && v[1]<=v[2] && v[2]<=v[3] && v[3]<=v[4])
                                                cout << "Yes\n";
                                else{
                                                cout << "No ";
                                                sort(v.begin(), v.end());
                                                vector<int>::iterator it;
                                                for(it = v.begin(); it != v.end(); it++)
                                                                cout << *it << " ";
                                                cout << endl;
                                }
                }
}
  
(2)、解题思路:略
(3)、可能出错:略
 
 
posted on 2012-05-11 20:12  谷堆旁边  阅读(412)  评论(0编辑  收藏  举报