Nine(2012 Rocky Mountain Regional Contest)

地址：http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11346&courseid=0

NineTime Limit: 1500ms, Special Time Limit:3000ms, Memory Limit:32768KBTotal submit users: 7, Accepted users: 6Problem 11346 : No special judgementProblem description   

Randall Munroe from xkcd.com pointed out that 9 is the most rarely used key on a microwave. Let's all share the load.

Given a desired cooking time, find a sequence of keys with the greatest number of 9's such that the resulting time has less than 10% error compared to the desired cooking time. In other words, if T is the desired cooking time in seconds, and T9 is the cooking time specified by the found sequence, then 10|T - T9| < T. If there are multiple possibilities, choose the one that has the smallest error (in magnitude). If there are still ties, choose the one that is lexicographically smallest.

For example, for T = 01:15, the times 00:68-00:82 and 1:08-1:22 have less than 10% error. Of these, 00:69, 00:79, 01:09, and 01:19 have the greatest number of 9's, and the ones with the smallest error are 00:79 and 01:19. The lexicographically smaller of these is 00:79.

Input  

The input consists of a number of cases. For each case, the desired cooking time in MM:SS format is specified on one line. Each M or S can be any digit from 0 to 9. The end of input is indicated by 00:00.

Output  

For each case, output on a single line the four keys to use as input to the microwave, in MM:SS format.

Sample Input

00:3001:0002:0091:3046:0300:00

Sample Output

00:2900:5901:5990:9949:99

题意：输入一个时间点，照在误差小于时间值的10%的一段时间内含9最多的时间点，如果遇到含9数目相同的则选与输入时间点的差值小的，如果差值一样，就选字典序小的

易错例子：09:34

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{int m,s;}ti[4];int textnine(node tk){int sum=0;if(tk.m/10==9) sum++;if(tk.m%10==9) sum++;if(tk.s/10==9) sum++;if(tk.s%10==9) sum++;return sum;}node changen(node tk){node ts;ts=tk;if(tk.s<40&&tk.m>0){ts.m--;ts.s+=60;}return ts;}int main(){node ans,gin;int i,j,m,max9,k,c;while(scanf("%d:%d",&gin.m,&gin.s)>0){if(!gin.m&&!gin.s) break;gin.m+=gin.s/60;gin.s%=60;max9=textnine(gin);ans=gin;c=0;ti[0]=changen(gin);k=textnine(ti[0]);if(k>max9){max9=k;ans=ti[0];}else if(k==max9){if(ans.m>ti[0].m){ans.m=ti[0].m;ans.s=ti[0].s;}else if(ans.m==ti[0].m&&ans.s>ti[0].s)ans.s=ti[0].s;}m=(gin.m*60+gin.s)/10;if(gin.s%10==0) m--;for(i=1;i<=m;i++){ti[0]=gin;ti[0].m+=(ti[0].s-i)/60;ti[0].s=(ti[0].s-i)%60;if(ti[0].s<0) {ti[0].m--;ti[0].s+=60;}ti[1]=changen(ti[0]);ti[2]=gin;ti[2].m+=(ti[2].s+i)/60;ti[2].s=(ti[2].s+i)%60;ti[3]=changen(ti[2]);for(j=0;j<4;j++){k=textnine(ti[j]);if(k>max9){max9=k;c=i;ans=ti[j];}else if(k==max9&&i==c){if(ans.m>ti[j].m) ans=ti[j];else if(ans.m==ti[j].m&&ans.s>ti[j].s)ans.s=ti[j].s;}}}printf("%02d:%02d\n",ans.m,ans.s);}return 0;}