思路:n在
Sk
中,二分找到
k
。
http://acm.hdu.edu.cn/showproblem.php?pid=1597
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define rep(i,a,b) for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x) memset(a,x,sizeof(a))
#define eps 1e-8
using namespace std;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;
typedef long long LL;
typedef unsigned long long ULL;
int T,m,k;
LL n;
int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};
void input() {
scanf("%lld",&n);
}
void solve() {
LL left = 1;
LL right = 100000;
LL mid ;
while(left < right) {
mid = (left + right) >> 1;
if(mid*(mid+1)/2 < n) left = mid + 1;
else right = mid ;
}
left --;
LL pos = n - left * (left + 1) / 2;
printf("%lld\n",(pos-1)%9+1);
}
int main(void) {
scanf("%d",&T); while(T--) {
input();
solve();
}
return 0;
}