HDU3944 DP? (LUCAS定理+阶乘预处理)

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此博客讨论了如何计算杨辉三角中从顶部到指定行和列的最小路径和，考虑了路径方向限制，并提供了求解算法及实现。

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Problem Description

Figure 1 shows the Yang Hui Triangle. We number the row from top to bottom 0,1,2,…and the column from left to right 0,1,2,….If using C(n,k) represents the number of row n, column k. The Yang Hui Triangle has a regular pattern as follows.
C(n,0)=C(n,n)=1 (n ≥ 0)
C(n,k)=C(n-1,k-1)+C(n-1,k) (0<k<n)
Write a program that calculates the minimum sum of numbers passed on a route that starts at the top and ends at row n, column k. Each step can go either straight down or diagonally down to the right like figure 2.
As the answer may be very large, you only need to output the answer mod p which is a prime.

Input

Input to the problem will consists of series of up to 100000 data sets. For each data there is a line contains three integers n, k(0<=k<=n<10^9) p(p<10^4 and p is a prime) . Input is terminated by end-of-file.

Output

For every test case, you should output "Case #C: " first, where C indicates the case number and starts at 1.Then output the minimum sum mod p.

Sample Input

1 1 2
4 2 7

Sample Output

Case #1: 0
Case #2: 5
题目分析：要求计算最小的和，
当 k < n/2 时需要将C(N,K) 化成 C(N,N-K);
原式 C(N,K)+C(N-1,K-1)+...+C(N-K,0)+K;
由 C(N-1,K)+C(N-1,K-1)=C(N,K) 原式可化为C(N+1，K)+K;
然后经过Lucas定理+对阶乘的预处理即可得出结果；
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 10005;

bool prime[N];

int p[N],f[N][N],inv[N][N],cnt,pth[N];

void isprime(){
    cnt=0;
    memset(prime,1,sizeof(prime));
    for(int i=2;i<N;i++){
        if(prime[i]){
            p[++cnt]=i;
            pth[i]=cnt;
            for(int j=i+i;j<N;j+=i)
                prime[j]=0;
        }
    }
}

int quick_mod(int a,int b,int m){
    int ans=1;
    a%=m;
    while(b){
        if(b&1){
            ans=ans*a%m;
            b--;
        }
        b>>=1;
        a=a*a%m;
    }
    return ans;
}

void init(){//预处理阶乘与逆元
    int i,j;
    for(int i=1;i<=cnt;i++){
        f[i][0]=inv[i][0]=1;
        for(int j=1;j<p[i];j++){
            f[i][j]=(f[i][j-1]*j)%p[i];
            inv[i][j]=quick_mod(f[i][j],p[i]-2,p[i]);
        }
    }
}

int com(int n,int m,int P){
    if(m>n) return 0;
    if(m==n) return 1;
    int t=pth[P];
    return f[t][n]*(inv[t][n-m]*inv[t][m]%P)%P;
}

int lucas(int n,int m,int P){
    if(m==0) return 1;
    return com(n%P,m%P,P)*lucas(n/P,m/P,P)%P;
}
int main()
{
    int cas=1,n,m,P;
    isprime();
    init();
    while(cin>>n>>m>>P){
        if(m<=n/2) m=n-m;
        n++;
        printf("Case #%d: %d\n",cas++,(m%P+lucas(n,m+1,P))%P);
    }
    return 0;
}