hdu_1036_Average is not Fast Enough_201311021335
Average is not Fast Enough!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3436 Accepted Submission(s): 1348
You have to process several teams. For each team you are given a list with the running times for every section of the race. You are to compute the average time per kilometer over the whole distance. That's easy, isn't it? So if you like the fun and challenge competing at this contest, perhaps you like a relay race, too. Students from Ulm participated e.g. at the "SOLA" relay in Zurich, Switzerland. For more information visit http://www.sola.asvz.ethz.ch/ after the contest is over.
#include <stdio.h>
#include <string.h> char s[][];
int time[]; int main()
{
int N,n;
double d;
int num,i,j;
double res1;
int res2;
scanf("%d",&N);
scanf("%lf",&d);
memset(s,,sizeof(s));
while(scanf("%3d",&num)!=EOF)
{
int t=,totle=,k=;
//char h,m1,m2,s1,s2;
for(i=;i<N;i++)
{
getchar();
scanf("%s",s[i]);
//printf("%s\n",s[i]);
//scanf("%c:%c%c:%c%c",&h,&m1,&m2,&s1,&s2);
t=(s[i][]-'')*+(s[i][]-'')*+(s[i][]-'')*+(s[i][]-'')*+s[i][]-'';
//t=(h-'0')*3600+((m1-'0')*10+(m2-'0'))*60+(s1-'0')*10+(s2-'0');
if(t>)
{
totle+=t;
}
else
{
k=;
continue;
}
}
if(k==)
{
printf("%3d: -\n",num);
}
else
{
res2 = (int)totle*1.0/(d*);
res1 = (totle*1.0/(d*) - res2)*;
if(res1>=59.5)
{res2+=;res1=;}
printf("%3d:%2d:%02.0lf min/km\n",num,res2,res1);
}
}
//while(1);
return ;
}
网上不错的解法:
#include<stdio.h>
int main()
{
int n;
double d;
int num;
char h,m1,m2,s1,s2;
scanf("%d",&n);
scanf("%lf",&d);
while(scanf("%d",&num)!=EOF)
{ printf("%3d: ",num);
bool flag=true;
int sumtime=;
for(int i=;i<n;i++)
{getchar();
scanf("%c:%c%c:%c%c",&h,&m1,&m2,&s1,&s2);
if(h=='-') flag=false;
if(flag==false)continue;
sumtime+=(h-'')*+((m1-'')*+(m2-''))*+(s1-'')*+(s2-'');
}
if(flag==false)printf("-\n");
else
{
double t1=sumtime/d;
int t2=(int)(t1+0.5);
printf("%d:%02d min/km\n",t2/,t2%); }
}
return ;
}
总结:题比较简单,需认真读题弄明白题意,注意输入的格式
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