1308. 方程的解(隔板法)

题目链接：https://www.acwing.com/activity/content/problem/content/1761/

 

本题需要用高精度

Code

#include 
#include 
#include 

using namespace std;

const int N = 150;

int k, x;
int f[1000][100][N];

int qmi(int a, int b, int p)
{
    int res = 1;
    while (b)
    {
        if (b & 1) res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}

void add(int c[], int a[], int b[])
{
    for (int i = 0, t = 0; i < N; i ++ )
    {
        t += a[i] + b[i];
        c[i] = t % 10;
        t /= 10;
    }
}

int main()
{
    cin >> k >> x;

    int n = qmi(x % 1000, x, 1000);

    // C(n - 1, k - 1)
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j <= i && j < k; j ++ )
            if (!j) f[i][j][0] = 1;
            else add(f[i][j], f[i - 1][j], f[i - 1][j - 1]);  // f[i][j] = f[i - 1][j] + f[i - 1][j - 1];

    int *g = f[n - 1][k - 1];
    int i = N - 1;
    while (!g[i]) i -- ;
    while (i >= 0) cout << g[i -- ];

    return 0;
}

如果要处理的是不等式，再加一个隔板即可，最后一个隔板隔开多余的球