Description：
Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that: f ( x ) = 5 ∗ x 13 + 13 ∗ x 5 + k ∗ a ∗ x f(x)=5*x^{13}+13*x^5+k*a*x f(x)=5∗x13+13∗x5+k∗a∗x,input a nonegative integer k ( k < 10000 ) k(k<10000) k(k<10000),to find the minimal nonegative integer a,make the arbitrary integer x x x , 65 ∣ f ( x ) 65|f(x) 65∣f(x)
if no exists that a,then print " n o " . "no". "no".
Input
The input contains several test cases. Each test case consists of a nonegative integer k k k, More details in the Sample Input.
Output
The output contains a string “ n o no no”,if you can’t find a,or you should output a line contains the a a a.More details in the Sample Output.
Sample Input
11
100
9999
Sample Output
22
no
43
题意：
给定一个方程式
f
(
x
)
=
5
∗
x
13
+
13
∗
x
5
+
k
∗
a
∗
x
f(x)=5*x^{13}+13*x^5+k*a*x
f(x)=5∗x13+13∗x5+k∗a∗x，给定一个非负整数k，求能不能找到一个尽量小的非负整数
a
a
a，使得上述方程式中的
x
x
x 任意取值，结果都能被
65
65
65 整除，如果有，输出
a
a
a 的值，否则输出
n
o
no
no。
我们采用数学归纳法，假设
a
a
a 对于
f
(
x
)
f(x)
f(x) 成立的话那么对于
f
(
x
+
1
)
f(x+1)
f(x+1) 也一定成立，
f
(
x
)
=
5
∗
x
13
+
13
∗
x
5
+
k
∗
a
∗
x
f(x)=5*x^{13}+13*x^5+k*a*x
f(x)=5∗x13+13∗x5+k∗a∗x
= 5 ∗ ( x + 0 ) 13 + 13 ∗ ( x + 0 ) 5 + k ∗ a ∗ ( x + 0 ) =5*(x+0)^{13}+13*(x+0)^5+k*a*(x+0) =5∗(x+0)13+13∗(x+0)5+k∗a∗(x+0)
f ( x + 1 ) = 5 ∗ ( x + 1 ) 13 + 13 ∗ ( x + 1 ) 5 + k ∗ a ∗ ( x + 1 ) f(x+1)=5*(x+1)^{13}+13*(x+1)^5+k*a*(x+1) f(x+1)=5∗(x+1)13+13∗(x+1)5+k∗a∗(x+1)
采用二项式定理， f ( x + 1 ) − f ( x ) = 18 + k ∗ a f(x+1)-f(x)=18 + k * a f(x+1)−f(x)=18+k∗a 如果是的要求成立的话那么 65 ∣ ( f ( x + 1 ) − f ( x ) ) 65|(f(x+1)-f(x)) 65∣(f(x+1)−f(x)) 即： 65 ∣ ( 18 + k ∗ a ) 65|(18 + k * a) 65∣(18+k∗a), 所以只要暴力求解就行了，我们输出一部分就会发现 a a a 都是小于 70 70 70 的
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
if (a >= mod)
a = a % mod + mod;
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = ans * a;
if (ans >= mod)
ans = ans % mod + mod;
}
a *= a;
if (a >= mod)
a = a % mod + mod;
b >>= 1;
}
return ans;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
int g = exgcd(b, a % b, x, y);
int t = x;
x = y;
y = t - a / b * y;
return g;
}
///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
int d, x, y;
d = exgcd(a, p, x, y);
if (d == 1)
return (x % p + p) % p;
else
return -1;
}
///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
int M = 1, y, x = 0;
for (int i = 0; i < n; ++i) //算出它们累乘的结果
M *= a[i];
for (int i = 0; i < n; ++i)
{
int w = M / a[i];
int tx = 0;
int t = exgcd(w, a[i], tx, y); //计算逆元
x = (x + w * (b[i] / t) * x) % M;
}
return (x + M) % M;
}
int main()
{
int k;
while (~sd(k))
{
int ans;
bool flag = 0;
rep(i, 0, 70)
{
if ((18 + k * i) % 65 == 0)
{
flag = 1;
ans = i;
break;
}
}
if (!flag)
puts("no");
else
pd(ans);
}
return 0;
}