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PAT 1006的坑

2013-11-10 17:05 302 查看
这道题要注意的陷阱就一点:如果客户A的账单记录没有任何一对是合法的,则客户A什么信息都不输出

解题思路是用一个结构体Record装每一条电话信息,按用户名字和时间排序。然后找匹配就可以了。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<math.h>

#undef DEBUG
using namespace std;

class Record
{
public:
char name[21];
int m,d,h,minute;
bool on;
void nextHour()
{
h++;
if(h==24)
{
h=0;
d++;
}
}
};

int rate[24];
int n;
vector<Record> v;

bool less_record(const Record & r1,const Record & r2)
{
if(strcmp(r1.name,r2.name) < 0)
return true;
if(strcmp(r1.name,r2.name) > 0)
return false;
if(r1.d<r2.d)
return true;
if(r1.d>r2.d)
return false;
if(r1.h<r2.h)
return true;
if(r1.h>r2.h)
return false;
if(r1.minute<r2.minute)
return true;
return false;
}

bool inSameHour(int i,int j)
{
if(v[i].d== v[j].d &&v[i].h== v[j].h)
return true;
else
return false;
}

int main()
{

int i;
for(i=0;i<24;i++)
scanf("%d",&rate[i]);
scanf("%d",&n);
for(i=0;i<n;i++)
{
Record r;
char s[21];
scanf("%s",r.name);
scanf("%d:%d:%d:%d",&r.m,&r.d,&r.h,&r.minute);
scanf("%s",s);
if(strcmp(s,"on-line")==0)
r.on=true;
else
r.on=false;
v.push_back(r);
}

sort(v.begin(),v.end(),less_record);

char nowName[21];
nowName[0]='\0';
int amount,totalAmount;
int talkTime;
totalAmount=-1;
bool hasLeft=false;
int leftIndex;
for(i=0;i<n;i++)
{
if(strcmp(nowName,v[i].name) !=0)
{
if(totalAmount>0)
{
printf("Total amount: $%.2lf\n",totalAmount/100.0);
}
strcpy(nowName,v[i].name);
totalAmount=0;
//printf("%s %02d\n",nowName,v[i].m);
hasLeft=false;
}
if(v[i].on)
{
hasLeft=true;
leftIndex=i;
}
else if(!hasLeft)
{
//do nothing
}
else
{
//have a pair
hasLeft=false;
if(totalAmount<=0)
printf("%s %02d\n",nowName,v[i].m);
printf("%02d:%02d:%02d %02d:%02d:%02d ",v[leftIndex].d,v[leftIndex].h,v[leftIndex].minute,v[i].d,v[i].h,v[i].minute);
if(inSameHour(leftIndex,i))
{
talkTime=v[i].minute-v[leftIndex].minute;
amount=talkTime*rate[v[i].h];
}
else
{
talkTime=60-v[leftIndex].minute;
amount=talkTime*rate[v[leftIndex].h];
v[leftIndex].minute=0;
v[leftIndex].nextHour();

talkTime+=v[i].minute;
amount+=v[i].minute*rate[v[i].h];
v[i].minute=0;

while(!inSameHour(leftIndex,i))
{
talkTime+=60;
amount+=60*rate[v[leftIndex].h];
v[leftIndex].nextHour();
}
}
totalAmount+=amount;
printf("%d $%.2lf\n",talkTime,amount/100.0);
}
}
if(totalAmount>0)
{
printf("Total amount: $%.2lf\n",totalAmount/100.0);
}
return 0;
}
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标签:  PAT