CodeForces 622A Infinite Sequence

简单题，公式打了个表，查询的时候二分一下就行。也可以直接o(1)公式出解。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <queue>
#include <stack>
#include <map>
#include <vector>
using namespace std;

long long a[20000005];

int main()
{
    for (int i = 1; i <= 20000000; i++)
    {
        long long now = (long long)i;
        a[i] = (1 + now)*now / 2;
    }

    long long n;
    while (~scanf("%lld", &n)){

        int l = 1, r = 20000000;
        int pos;
        while (l <= r)
        {
            int mid = (l + r) / 2;
            if (a[mid] <= n) { pos = mid; l = mid + 1; }
            else r = mid - 1;
        }

        if (a[pos] == n) printf("%d\n", pos);
        else printf("%lld\n", n - a[pos]);
    }
    return 0;
}