三角形数且是完全平方数 2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 E.Half-consecutive Numbers

三角形数：an=n*(n+1)/2;

完全平方数：bn=c^2;

既是三角形数又是完全平方数：An=6*A(n-1)-A(n-2)+2;

A[23]={ 0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761,511643454094368, 2982076586042449, 17380816062160328};

所以这道题只需要打表就好了。。

注意是让r>=N,a存的是下标。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll a[23] = { 0, 1, 8, 49, 288, 1681, 9800, 57121, 332928,1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048,
76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761,511643454094368, 2982076586042449, 17380816062160328 };

int main()
{
    int T,n;
    cin >> T;
    for (int cas = 1; cas <= T; cas++) {
        cin >> n;
        for (int i = 0; i < 23; i++) {
            if (a[i] >= n) {
                cout << "Case #" << cas << ": " << a[i] << endl;
                break;
            }
        }
    }
    return 0;
}

 还有个脑洞特大的做法：

http://blog.csdn.net/hyesuixin/article/details/77916761

其他：

http://blog.csdn.net/a15110103117/article/details/77916895

(感谢博主）