PAT 1006. Sign In and Sign Out
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如果一个人的离开时间比另一个人的到达时间还要早呢?
不过这道题目好像不用这么复杂考虑。
简单的字符串输入输出问题。
直接把时间换成已秒为单位去求解就好了。
以上。
#include<stdio.h>#include<math.h>#include<algorithm>#include<stdlib.h>#include<iostream>#include<string.h>#include<queue>using namespace std;#define maxn 505char id[maxn][20];int sign_in[maxn][3];int sign_out[maxn][3];int in[maxn],out[maxn];int time(int i,int j,int k){ return i*3600+j*60+k;}int main(){ int m,i,j,k; scanf("%d",&m); for(i=0;i<m;i++) { scanf("%s",&id[i]); scanf("%d:%d:%d",&sign_in[i][0],&sign_in[i][1],&sign_in[i][2]); in[i]=time(sign_in[i][0],sign_in[i][1],sign_in[i][2]); scanf("%d:%d:%d",&sign_out[i][0],&sign_out[i][1],&sign_out[i][2]); out[i]=time(sign_out[i][0],sign_out[i][1],sign_out[i][2]); } int minn=86400; for(i=0;i<m;i++) { if(in[i]<minn) { minn=in[i]; j=i; } } int maxx=0; for(i=0;i<m;i++) { if(out[i]>maxx) { maxx=out[i]; k=i; } } printf("%s %s",id[j],id[k]); return 0;}
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