标签：deb using name 高精度 private highlight 进一步 一个个 struct
对于不定方程a1+a2+…+ak-1+ak=g(x)，其中k≥2且k∈N，x是正整数，g(x)=x^x mod 1000(即x^x除以1000的余数)，x，k是给定的数。我们要求的是这个不定方程的正整数解组数。
定义F(p, rest)为第p个数，p及p后面的数的和为rest的解的数量，递归式为：F(p, rest)=if(rest==0)0 else if (p == k) rest else sum foreach curVal(0<curVal<=rest) F(p, rest - curVal)。
不用高精度，本方法能得40分。
curVal<=rest就可以了，反正如果curVal大了以后也能返回0，不要擅自进一步缩小curVal的范围。
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
#define _NDEBUG
const int MAX_K = 110, MAX_X = 1010, P = 1000;
ll F[MAX_K][MAX_X];
ll K, X;
ll Mult(ll a, ll b, ll p)
{
ll ans = 0;
while (b)
{
if (b & 1)
ans = (ans + a) % p;
a = (a + a) % p;
b >>= 1;
}
return ans;
}
ll Power(int a, int n, int p)
{
ll ans = 1;
while (n)
{
if (n & 1)
ans = Mult(ans, a, p);
a = Mult(a, a, p);
n >>= 1;
}
return ans;
}
ll DP(int p, ll rest)
{
if (F[p][rest] != -1)
return F[p][rest];
if (rest == 0)
return F[p][rest] = 0;
if (p == K)
return F[p][rest] = 1;
F[p][rest] = 0;
for (int i = 1; i <= rest; i++)
F[p][rest] += DP(p + 1, rest - i);
return F[p][rest];
}
int main()
{
scanf("%lld%lld", &K, &X);
memset(F, -1, sizeof(F));
printf("%lld\n", DP(1, Power(X, X, P)));
return 0;
}
把x^x%1000看作一个线段，a1,a2...看作一个个隔板划分出的区域，这就是组合数学中的隔板模型，最终结果为C(k,x^x%1000)
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
const int MAX_K = 110, MAX_X = 1010, P = 1000, BASE = 10000, CARRY = 4, MAX_LEN = 1010;
ll K, X;
ll Mult(ll a, ll b, ll p)
{
ll ans = 0;
while (b)
{
if (b & 1)
ans = (ans + a) % p;
a = (a + a) % p;
b >>= 1;
}
return ans;
}
ll Power(int a, int n, int p)
{
ll ans = 1;
while (n)
{
if (n & 1)
ans = Mult(ans, a, p);
a = Mult(a, a, p);
n >>= 1;
}
return ans;
}
struct BigInt
{
private:
int A[MAX_LEN];
int Len;
public:
void Clear()
{
memset(A, 0, sizeof(A));
Len = 0;
}
void Set(int x)
{
Clear();
while (x)
{
A[Len++] = x%BASE;
x /= BASE;
}
while (A[Len] == 0 && Len > 0)
Len--;
}
BigInt() { Clear(); }
BigInt operator = (const int x)
{
Set(x);
return *this;
}
BigInt operator = (const BigInt &a)
{
memcpy(A, a.A, sizeof(A));
Len = a.Len;
return *this;
}
void Print()
{
printf("%d", A[Len]);
for (int i = Len - 1; i >= 0; i--)
printf("%0*d", CARRY, A[i]);
printf("\n");
}
BigInt operator += (const BigInt &a)
{
Len = max(Len, a.Len);
for (int i = 0; i <= Len; i++)
{
A[i] += a.A[i];
A[i + 1] += A[i] / BASE;
A[i] %= BASE;
}
if (A[Len + 1])
Len++;
return *this;
}
BigInt operator + (const BigInt &a)
{
BigInt Num = *this;
return Num += a;
}
};
BigInt Comb(int r, int n)
{
static BigInt C[MAX_K];
C[0] = 1;
for (int i = 1; i <= n; i++)
{
for (int j = min(r, i); j > 0; j--)
{
if (i == j)
C[j] = 1;
else
C[j] = C[j - 1] + C[j];
}
}
return C[r];
}
int main()
{
scanf("%lld%lld", &K, &X);
Comb(K - 1, Power(X,X,P) - 1).Print();
return 0;
}
标签：deb using name 高精度 private highlight 进一步 一个个 struct
原文地址：https://www.cnblogs.com/headboy2002/p/8886493.html
