http://codeforces.com/contest/716/problem/C
题目大意：感觉这道题还是好懂得吧。
思路：不断的通过列式子的出来了。首先我们定义level=i, uplevel = i + 1，目前的uplevel，然后我们可以知道，之前求出来的restgrade%level = 0,于是我们可以列出一个式子sqrt(k1 * i + grade) = uplevel * k.因为grade也是i的倍数,假定grade = k2 * i，所以我们写成sqrt((k1 + k2) * i) = uplevel * k,所以两边平方再移项以后就可以发现总有这样的一个解满足条件，即:k1 = level * uplevel * uplevel - k2。然后我们打表一下：
//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n") int main(){
LL n;
LL level = , grade = ;
LL k1, k2, k;
scanf("%I64d", &n);
for (LL uplevel = ; uplevel <= n + ; uplevel++){
k2 = grade / level;
///printf("k2 = %I64d\n", k2);
k1 = level * uplevel * uplevel - k2;
grade = sqrt((k1 + k2) * level);
///printf("grade = %I64d %I64d %I64d\n", grade, grade / level , uplevel);
level++;
printf("%I64d\n", k1);
}
return ;
}
我们发现grade总等于level * uplevel，所以我们就可以求得level，然后看到level=1的时候不满足条件，然后特判一下就好了
//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n") int main(){
LL n;
LL level = , grade = ;
LL k1, k2, k;
scanf("%I64d", &n);
cout << << endl;
for (LL i = ; i <= n; i++){
printf("%I64d\n", (i + ) * (i + ) * i - i + );
}
return ;
}