Codeforces Round #372 (Div. 2) C 数学

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
")


int main(){
    LL n;
    LL level = 1, grade = 2;
    LL k1, k2, k;
    scanf("%I64d", &n);
    for (LL uplevel = 2; uplevel <= n + 1; uplevel++){
        k2 = grade / level;
        ///printf("k2 = %I64d
", k2);
        k1 = level * uplevel * uplevel - k2;
        grade = sqrt((k1 + k2) * level);
        ///printf("grade = %I64d %I64d %I64d
", grade, grade / level , uplevel);
        level++;
        printf("%I64d
", k1);
    }
    return 0;
}

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
")


int main(){
    LL n;
    LL level = 1, grade = 2;
    LL k1, k2, k;
    scanf("%I64d", &n);
    cout << 2 << endl;
    for (LL i = 2; i <= n; i++){
        printf("%I64d
", (i + 1) * (i + 1) * i - i + 1);
    }
    return 0;
}