九度互动社区IT名企招聘上机考试热身赛

http://ac.jobdu.com/problem.php?pid=1326     Waiting in Line

//简单模拟题#include<iostream>#include<cstdio>using namespace std;#include<memory.h>int pt[1001],leave[1001],start[21];int n,m,k,q;inline bool scan_d(int &num)  //  这个就是 加速的 关键了   {char in;bool IsN=false;in=getchar();if(in==EOF)return false;while(in!='-'&&(in<'0'||in>'9')) in=getchar();if(in=='-')   { IsN=true;num=0;}else num=in-'0';while(in=getchar(),in>='0'&&in<='9'){num*=10,num+=in-'0';}if(IsN)num=-num;return true;}void solve(){int i,j,index,t,p,count=k;int num[21],window[21][13];memset(start,0,sizeof(start));memset(num,0,sizeof(num));j = index = 1;for(i = 1 ; i <= n*m && i <= k ; ++i)    //刚开始时，n个窗口都是空的，客户依次选择队伍最短的窗口{window[j][index] = i;num[j]++;j++;if(j>n){j = 1;++index;}}j = i;   //刚开始的n个窗口已经都排满了客户t = 0;while(t <= 540){++t;for(i = 1 ; i <= n ; ++i){if(num[i] > 0 && pt[ window[i][1] ] == t - start[i]){leave[ window[i][1] ] = t;start[i] = t;--count;        //一个用户离开for(p = 2 ; p <= num[i] ; ++p)window[i][p-1] = window[i][p];    //每个窗口的第一个用户离开后，后面的客户依次往前移动--num[i];if(j <= k)    //还有等待的客户的时候，往窗口中再增加一个{window[i][p-1] = j;++num[i];++j;}}//if}//forif(!count)    //所有客户都处理完的时候，直接退出break;}}int main(void){int i,j;while(scanf("%d %d %d %d",&n,&m,&k,&q)!=EOF){memset(leave,1,sizeof(leave));for(i = 1 ; i <= k ; ++i)scan_d(pt[i]);solve();for(i = 0 ; i < q ; ++i){scan_d(j);if(leave[j] > 540)puts("Sorry");elseprintf("%02d:%02d\n",leave[j]/60+8,leave[j]%60);}}return 0;}