hdu 1868

简单的找规律题，假设连续序列为a, a+1, a+2,...,a+k则(2*a+k)*(k+1)/2=N，从而a可以用N和k表示出来，而通过这个公式能够发现k一定小于sqrt(2*N+1)，枚举k，计算出a判断是否符合题意即可。

/*
 * hdu1868/win.cpp
 * Created on: 2011-10-8
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int N, end, itmp;
    double dtmp;
    int ans;
    while (scanf("%d", &N) == 1) {
        ans = 0;
        end = (int) sqrt(2 * N + 1.0);
        for (int k = 1; k < end; k++) {
            dtmp = (N - k * k / 2.0 - k / 2.0) / (k + 1.0);
            itmp = (int) dtmp;
            if (itmp == dtmp && itmp >= 1 && itmp < N) {
                ans++;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}