锘??xml version="1.0" encoding="utf-8" standalone="yes"?>午夜免费电影一区在线观看,国产精品护士白丝一区av,日韩视频在线你懂得http://m.shnenglu.com/zhuimengboy65/category/8588.html浠庡悙涓濈粨鑼у埌鐮磋導鑰屽嚭zh-cnSat, 29 May 2010 21:04:18 GMTSat, 29 May 2010 21:04:18 GMT60鎶婅嚜1970騫?鏈?鏃ヤ互鏉ョ殑縐掓暟杞寲鎴愬勾鏈堟棩http://m.shnenglu.com/zhuimengboy65/archive/2010/05/29/116684.html鏄ュぉ鏄ュぉSat, 29 May 2010 05:54:00 GMThttp://m.shnenglu.com/zhuimengboy65/archive/2010/05/29/116684.htmlhttp://m.shnenglu.com/zhuimengboy65/comments/116684.htmlhttp://m.shnenglu.com/zhuimengboy65/archive/2010/05/29/116684.html#Feedback0http://m.shnenglu.com/zhuimengboy65/comments/commentRss/116684.htmlhttp://m.shnenglu.com/zhuimengboy65/services/trackbacks/116684.html#include <iostream>
using namespace std;

bool IsRound(int year){
 if((year%100)&&(year%4==0)) return 1;
 if((year%100==0)&&(year%400==0)) return 1;
 return 0;
}
int main() {
 int n;
 int year_s[2]= {365*24*60*60, 366*24*60*60};
 int month_s[2][12]={{31,28,31,30,31,30,31,31,30,31,30,31},
   {31,29,31,30,31,30,31,31,30,31,30,31}};
 int day_s = 24*60*60;
 int hour_s = 60*60;
 int minute_s = 60;

 

 while(cin>>n){
    int temp = n;
    int year=1970;
         int month=1;
         int day=1;
         int hour=0;
         int minute=0;
         int second=0;

    while(temp>=60){

     int flag= IsRound(year);
     if(temp>=year_s[flag]) { year++; temp-=year_s[flag]; }
     else if(temp>=day_s){
      int days = temp/day_s;
      temp=temp%day_s;
      int i=0;
      int flag = IsRound(year);
      int hh=31;
      while(days>=hh){
       days-=month_s[flag][i++];
       hh=month_s[flag][i];
       }
      month+= i;
      day+=days;

     }else if(temp>=hour_s){
          hour=temp/hour_s;
          temp%=hour_s;
     }else if(temp>=minute_s){
           minute = temp/minute_s;
           temp%=minute_s;
     }
     }
     second = temp;

        cout<<year<<"-";
      if(month/10==0) cout<<"0";
      cout<<month<<"-";
      if(day/10==0) cout<<"0";
      cout<<day<<" ";
      if(hour/10==0) cout<<"0";
      cout<<hour<<":";
      if(minute/10==0) cout<<"0";
      cout<<minute<<":";
      if(second/10==0) cout<<"0";
      cout<<second<<endl;
 }


 return 0;
}



鏄ュぉ 2010-05-29 13:54 鍙戣〃璇勮
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DFS,BFS,DFS+IDhttp://m.shnenglu.com/zhuimengboy65/archive/2008/10/22/64723.html鏄ュぉ鏄ュぉWed, 22 Oct 2008 11:02:00 GMThttp://m.shnenglu.com/zhuimengboy65/archive/2008/10/22/64723.htmlhttp://m.shnenglu.com/zhuimengboy65/comments/64723.htmlhttp://m.shnenglu.com/zhuimengboy65/archive/2008/10/22/64723.html#Feedback0http://m.shnenglu.com/zhuimengboy65/comments/commentRss/64723.htmlhttp://m.shnenglu.com/zhuimengboy65/services/trackbacks/64723.html    DFS O(c k) O(k) Must search tree anyway, know the level the answers are on, or you aren't looking for the shallowest number.
    BFS O(c d ) O(c d ) Know answers are very near top of tree, or want shallowest answer.
    DFS+ID O(c d) O(d) Want to do BFS, don't have enough space, and can spare the time.
d is the depth of the answer k is the depth searched d <= k Remember the ordering properties of each search. If the program needs to produce a list sorted shortest solution first (in terms of distance from the root node), use breadth first search or iterative deepening. For other orders, depth first search is the right strategy. If there isn't enough time to search the entire tree, use the algorithm that is more likely to find the answer. If the answer is expected to be in one of the rows of nodes closest to the root, use breadth first search or iterative deepening. Conversely, if the answer is expected to be in the leaves, use the simpler depth first search. Be sure to keep space constraints in mind. If memory is insufficient to maintain the queue for breadth first search but time is available, use iterative deepening.
    quote from http://ace.delos.com/usacotext2?a=y9SZdbB6WeB&S=rec

鏄ュぉ 2008-10-22 19:02 鍙戣〃璇勮
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