URAL 1209 - 1, 10, 100, 1000... (简单推导)

题意

判断第N个是不是1

思路

发现在$1 + (1 + x) * x / 2$上的都是1,然后只要判断这个位置即可。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 10000 + 10;
const int MOD = 1e9 + 7;
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
int main()
{
    int T, i, j, n;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        n--;
        LL k = 2ll * n;
        LL l = (int)(sqrt(1 + 4 * k) / 2);
        LL r = l + 1;
        if ((1 + l) * l == k || (1 + r) * r == k) printf("1 ");
        else printf("0 ");
        puts("");
    }
 
}

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