HDOJ 1201 18岁生日

18岁生日

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 23040 Accepted Submission(s): 7351



[align=left]Problem Description[/align]
Gardon的18岁生日就要到了，他当然很开心，可是他突然想到一个问题，是不是每个人从出生开始，到达18岁生日时所经过的天数都是一样的呢？似乎并不全都是这样，所以他想请你帮忙计算一下他和他的几个朋友从出生到达18岁生日所经过的总天数，让他好来比较一下。

[align=left]Input[/align]
一个数T，后面T行每行有一个日期，格式是YYYY-MM-DD。如我的生日是1988-03-07。

[align=left]Output[/align]
T行，每行一个数，表示此人从出生到18岁生日所经过的天数。如果这个人没有18岁生日，就输出-1。

[align=left]Sample Input[/align]

1
1988-03-07

[align=left]Sample Output[/align]

6574

[align=left]Author[/align]
Gardon

[align=left]Source[/align]
Gardon-DYGG Contest 2

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注意 if(yun(a)&&b==2&&c==29) return -1;

#include<stdio.h>
#include<string.h>
struct birth{
int year,mon,day;
}bir[200];
int yun(int a){
if(a%400==0||(a%100!=0&&a%4==0))
return 1;
else return 0;
}
int dday(int a,int b,int c){
int sum=0,i,starsum=0,endsum=0;
if(yun(a)&&b==2&&c==29) return -1;
for(i=(a+1);i<=(a+17);i++){
if(yun(i)) sum+=366;
else sum+=365;
}
if(yun(a)){
switch(b){
case 1: starsum+=31;
case 2: starsum+=29;
case 3: starsum+=31;
case 4: starsum+=30;
case 5: starsum+=31;
case 6: starsum+=30;
case 7: starsum+=31;
case 8: starsum+=31;
case 9: starsum+=30;
case 10: starsum+=31;
case 11: starsum+=30;
case 12: starsum+=31; break;
}
starsum-=c;
}
else{
switch(b){
case 1: starsum+=31;
case 2: starsum+=28;
case 3: starsum+=31;
case 4: starsum+=30;
case 5: starsum+=31;
case 6: starsum+=30;
case 7: starsum+=31;
case 8: starsum+=31;
case 9: starsum+=30;
case 10: starsum+=31;
case 11: starsum+=30;
case 12: starsum+=31; break;
}
starsum-=c;
}
if(yun(a+18)){
switch(b){
case 1: endsum+=31;
case 2: endsum+=29;
case 3: endsum+=31;
case 4: endsum+=30;
case 5: endsum+=31;
case 6: endsum+=30;
case 7: endsum+=31;
case 8: endsum+=31;
case 9: endsum+=30;
case 10: endsum+=31;
case 11: endsum+=30;
case 12: endsum+=31; break;
}
endsum=(366-endsum+c);
}
else {
switch(b){
case 1: endsum+=31;
case 2: endsum+=28;
case 3: endsum+=31;
case 4: endsum+=30;
case 5: endsum+=31;
case 6: endsum+=30;
case 7: endsum+=31;
case 8: endsum+=31;
case 9: endsum+=30;
case 10: endsum+=31;
case 11: endsum+=30;
case 12: endsum+=31; break;
}
endsum=(365-endsum+c);
}
sum+=(starsum+endsum);
return sum;
}
int main(){
int n,i;
scanf("%d",&n);
for(i=0;i<n;i++){
memset(bir,0,sizeof(bir));
scanf("%d-%d-%d",&bir[i].year,&bir[i].mon,&bir[i].day);
printf("%d\n",dday(bir[i].year,bir[i].mon,bir[i].day));
}
return 0;
}

牛人代码（计算18年每年比365多几天，再加上）：

<span style="color:#000000;">#include "stdio.h"
int main()
{int n,i,y,m,d,sum,x;
scanf("%d",&n);
while(n--)
{   sum=0;//18年每年比365多几天
scanf("%d-%d-%d",&y,&m,&d);
if(m==2&&d==29)
{
printf("-1\n");
continue;
}
if((y%4==0&&y%100!=0||y%400==0)&&m<3)
sum++;
x=y+18;
if((x%4==0&&x%100!=0||x%400==0)&&m>=3)
sum++;
for(i=y+1;i<y+18;i++)
if(i%4==0&&i%100!=0||i%400==0)
sum++;
printf("%d\n",365*18+sum);
}
return 0;
}</span>