Hdoj 1201-18 Birthday

18-year-old birthday

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 23314 Accepted Submission (s): 7461

Problem Descriptiongardon's 18 birthday is coming, he is certainly very happy, but he suddenly thought of a question, is everyone from birth, to reach the 18 birthday when the number of days are the same? It doesn't seem to be all that, so he wants to ask you to calculate the total number of days he and his friends have passed since they were born to their 18 birthday.

Input is a number T, and the following T line has a date on each line, and the format is YYYY-MM-DD. As my birthday is in 1988-03-07.

Outputt lines, one number per line, indicating the number of days that the person has passed from birth to the age of 18. If this person does not have a 18 birthday, then output-1.

Sample INPUT1 1988-03-07

Sample Output65741, pay attention to judge leap Year 2, when the birthday is 02 29 o'clock to determine whether the age of 18 is a leap years if it is not a leap-year, there is no 18 birthday

#include <stdio.h> #include <string.h>int run (int x) {if (x%4==0&&x%100!=0| |    x%400==0) return 1; return 0;}    int main () {int n,j,i,s,t,l,sum;    int y,m,d;    int s1[14]={0,31,28,31,30,31,30,31,31,30,31,30,31};    int s2[14]={0,31,29,31,30,31,30,31,31,30,31,30,31};    scanf ("%d", &n);        while (n--) {scanf ("%d-%d-%d", &y,&m,&d);                if (!run (y+18)) {if (m==2&&d==29) {printf (" -1\n");            Continue        }} s=0;            for (i=y;i<y+18;i++) {if (run (i)) s+=366;        else s+=365;            } if (run (y+18)) {//s+=366;            for (i=1;i<m;i++) {s+=s2[i];        } S+=d;            } else {//s+=365;            for (i=1;i<m;i++) {s+=s1[i];        } S+=d;     } sum=0;   if (run (y)) {for (i=1;i<m;i++) {sum+=s2[i];        } Sum+=d;            } else {for (i=1;i<m;i++) {sum+=s1[i];        } Sum+=d;    } printf ("%d\n", s-sum); } return 0;}

Hdoj 1201-18 Birthday