Problem Description
Ignatius was born in a leap year, so he want to know when he could hold his birthday party. Can you tell him?
Given a positive integers Y which indicate the start year, and a positive integer N, your task is to tell the Nth leap year from year Y.
Note:
if year Y is a leap year, then the 1st leap year is year Y.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains two positive integers Y and N(1<=N<=10000).
Output
For each test case, you should output the Nth leap year from year Y.
Sample Input
3
2005 25
1855 12
2004 10000
Sample Output
2108
1904
43236
/* 该题N定义为50000RE,100000AC 因为该题year没范围,但因为要可以求第10000个生日, 所以 year的范围也绝对不会是INT_MAX, 考虑到不知year的范围,先试用is_leap(int year) 函数判断 闰年,不用初始化,若超时,再按我这种方法 */ #include<string.h> #include<stdlib.h> #define N 100000 int leap[N]; void isleap() { int i,j; memset(leap,0,sizeof(leap)); for(i=1;i<N;i++) if(i%4==0&&i%100!=0||i%400==0) leap[i]=1; } int main() { int i,j,T; int year,num,cnt; scanf("%d",&T); isleap(); while(T--) { cnt=0; scanf("%d %d",&year,&num); for(i=year;cnt<num;i++) if(leap[i]) cnt++; printf("%d\n",i-1); } system("pause"); return 0; }