題目不難,但輸入輸出數據多,格式多,需要些許思維和注意。花費了非常多時間,也算有些收穫。首先,卡在輸入輸出上,string類型只能用cin,不能用scanf,printf ,若要用printf,只能用printf("%s",str.c_str())。輸入中若帶“:”,得用getchar()。在printf時,最後一個%02d少加了個%,一直報越界。要細心。此題發現,pat測試數據愛出在臨界條件上,如01:02:14:58-01:02:14:59 ,若像我原本只考慮時間加頭加尾加中間的,就會出錯。另外,值得注意的點是,It is guaranteed that at least one call is well paired in the input. 只保證每次輸入都至少有一次正確匹配,不保證每個用戶都有一次正確匹配,因此可能出現有用戶無一次正確匹配,此時不要輸出用戶的任何信息。我還是太弱了~
AC代碼:
//1016 13:00
#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
const int NUM=1005;
struct record
{
string name;
int dd,hh,mm;
string status;
int total;
}rec[NUM];
struct call
{
int s;
int e;
double m;
}cal[NUM];
int toll[26];
int sumToll;
int cmp(const void *a, const void *b)
{
record *x=(record *)a;
record *y=(record *)b;
// if(x->name==y->name&&x->dd==y->dd&&x->hh==y->hh)
// return x->mm - y->mm;
// if(x->name==y->name&&x->dd==y->dd)
// return x->hh - y->hh;
// if(x->name==y->name)
// return x->dd - y->dd;
// return x->name < y->name;
if(x->name==y->name)
return x->total - y->total;
if(x->name < y->name)
return -1;
else return 1;
}
int find(int start,int end)
{
int ans=0,i;
if(rec[start].dd<rec[end].dd){
for(i=rec[start].hh+1;i<24;i++){
ans+=toll[i]*60;
}
for(i=0;i<rec[end].hh;i++)
ans+=toll[i]*60;
ans+=(60-rec[start].mm)*toll[rec[start].hh];
ans+=rec[end].mm*toll[rec[end].hh];
ans+=(rec[end].dd-rec[start].dd-1)*sumToll*60;
} else if(rec[start].dd==rec[end].dd) {
if(rec[end].hh>rec[start].hh){
for(i=rec[start].hh+1;i<rec[end].hh;i++)
ans+=toll[i]*60;
ans+=(60-rec[start].mm)*toll[rec[start].hh];
ans+=rec[end].mm*toll[rec[end].hh];
} else {
ans+=(rec[end].mm-rec[start].mm)*toll[rec[end].hh];
}
} else
ans=-1;
return ans;
}
int main()
{
int n,i,j,month;
freopen("C:\\Documents and Settings\\Administrator\\桌面\\input.txt","r",stdin);
sumToll=0;
for(i=0;i<24;i++){
cin>>toll[i];
sumToll+=toll[i];
}
cin>>n;
for(i=0;i<n;i++){//string類型只能用cin,不能用scanf,printf ,若要用printf,只能用printf("%s",str.c_str())
//cin>>rec[i].name>>month>>":">>rec[i].dd>>":">>rec[i].hh>>":">>rec[i].mm>>rec[i].status;
cin>>rec[i].name;
cin>>month;
getchar();
cin>>rec[i].dd;
getchar();
cin>>rec[i].hh;
getchar();
cin>>rec[i].mm;
cin>>rec[i].status;
rec[i].total=rec[i].dd*60*24+rec[i].hh*60+rec[i].mm;
}
qsort(rec,n,sizeof(rec[0]),cmp);
string str="";
int start,end,flag=0;
int theSame=0;
rec[n].name="";
double money=0.0f;
int minute;
double sum=0.0f;
int index=0;
for(i=0;i<n;i++){
if(str!=rec[i].name){
str=rec[i].name;
flag=0;
theSame=0;
index=0;
} else
theSame=1;//與前一個相同
if(rec[i].status[1]=='n'){//on
start=i;
flag=1;//1表示前面已經出現on了
}
if(theSame==1&&flag==1&&rec[i].status[1]=='f'){//off 和前面是同一個
end=i;
flag=0;//0 表示出現off了,結束了,可以重新開始
money=find(start,end);
money=money/100.0;
sum+=money;
cal[index].s=start;
cal[index].e=end;
cal[index].m=money;
//printf("%02d:%02d:%02d %02d:%02d:%02d %d $%0.2lf\n",rec[start].dd,rec[start].hh,rec[start].mm,rec[end].dd,rec[end].hh,rec[end].mm,minute,money);
index++;
}
if(rec[i].name!=rec[i+1].name&&index!=0){
printf("%s %02d\n",rec[i].name.c_str(),month);
for(j=0;j<index;j++){
start=cal[j].s;
end=cal[j].e;
money=cal[j].m;
minute=rec[end].total-rec[start].total;
printf("%02d:%02d:%02d %02d:%02d:%02d %d $%0.2lf\n",rec[start].dd,rec[start].hh,rec[start].mm,rec[end].dd,rec[end].hh,rec[end].mm,minute,money);
}
printf("Total amount: $%.2lf\n",sum);
sum=0.0f;
index=0;
}
}
// printf("\n");
// for(i=0;i<n;i++){
// printf("%s %02d:%02d:%02d:%02d %s\n",rec[i].name.c_str(),month,rec[i].dd,rec[i].hh,rec[i].mm,rec[i].status.c_str());
// }
return 0;
}