小米 oj 22 题 找到第 N 个数字

思路：二分找个位置，然后按循环节找就行。

#include 
#define LL long long 
using namespace std;
LL n ; 
LL cal( LL a ){
	return 1LL * a * ( a + 1 ) / 2 ; 
}
char s[] = "1234567898765432" ; 
int main(){
	scanf("%lld",&n) ; 
	LL l = 0 , r = 10000000000 ; 
	while( l <= r ){
		LL mid = ( l + r ) >> 1 ;
		if( cal(mid) >= n ){
			r = mid - 1 ;
		}else l = mid + 1 ; 
	}
	LL tmp = cal(l) ; 
	while( tmp >= n ){
		l -- ;
		tmp = cal(l) ; 
	}  
	n -= tmp ;
	n -- ; 
	printf("%c\n",s[n%16]);
	return 0 ; 
}