首先明確一點：最優解必為奶牛1..n-1輪流領跑，奶牛n撞線。且跑了x圈后，未領跑過的奶牛都耗費了x的體力。
設f[i][j][k]表示前i-1頭奶牛已領跑，現在由第i頭奶牛領跑，一共跑了j圈，奶牛i耗費了k的體力。
則f[i][j][k]可以轉移到f[i][j + p][k + p2]（耗費1分鐘，奶牛i以p圈/分鐘的速度繼續領跑），也可轉移到f[i + 1][j][j]（換成奶牛i + 1領跑，不耗費時間）。
時間復雜度為O(nde2.5)。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-9-1 10:45:17
File Name: pku1946.cpp
Description:
************************************************************************/
#include <iostream>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
const int maxint = 0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) {for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
int n, d, e;
int f[22][101][101];
int main()
{
scanf("%d%d%d", &n, &e, &d);
for (int i = 0; i <= n; i++)
for (int j = 0; j <= d; j++)
for (int k = 0; k <= e; k++)
f[i][j][k] = maxint;
f[1][0][0] = 0;
for (int i = 1; i <= n; i++)
for (int j = 0; j <= d; j++)
for (int k = 0; k <= e; k++) if (f[i][j][k] < maxint)
{
for (int p = 1; k + p * p <= e; p++)
f[i][j + p][k + p * p] <?= f[i][j][k] + 1;
f[i + 1][j][j] <?= f[i][j][k];
}
int ans = maxint;
for (int j = 0; j <= e; j++)
ans <?= f[n][d][j];
if (ans == maxint) printf("0\n");
else printf("%d\n", ans);
return 0;
}