hdu 3944 DP? lucas定理

DP?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 128000/128000 K (Java/Others)

Total Submission(s): 2460 Accepted Submission(s): 769



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

Author

phyxnj@UESTC



Source

2011 Multi-University Training Contest 11 -
Host by UESTC
预处理出所有质数p，n!%p以及n！对于p的逆元，
然后 http://www.tuicool.com/articles/E3yaIn看这个博客吧。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 10001
int check[maxn];
int jie[maxn][maxn];
int ni[maxn][maxn];
int cal(int n,int t,int p){
    int ans = 1;
    while(t){
        if(t&1) ans = ans*n%p;
        n = n*n%p;
        t/=2;
    }
    return ans;
}

void init(){
    memset(check,0,sizeof(check));
    for(int i = 2;i < maxn; i++){
        if(!check[i]){
            for(int j=i+i;j < maxn; j+=i)
                check[j] = 1;
            jie[i][0] = ni[i][0] = 1;
            for(int j = 1;j < i ;j++){
                jie[i][j] = jie[i][j-1]*j%i;
                ni[i][j] = cal(jie[i][j],i-2,i);
            }
        }
    }
}
int Lucas(int n,int m,int p){
    if(m == 0) return 1;
    int ans = 1;
    int a = n%p,b=m%p;
    if(b > a) return 0;
    ans = jie[p][a]*ni[p][a-b]%p*ni[p][b]%p;
    return Lucas(n/p,m/p,p)*ans%p;
}
int main(){
    init();
    int tt=1,n,m,p,ans;
    while(scanf("%d%d%d",&n,&m,&p)!=EOF){
        if(n/2>=m) ans = (Lucas(n+1,m,p)+n-m)%p;
        else ans = (Lucas(n+1,m+1,p)+m)%p;
        printf("Case #%d: %d\n",tt++,ans);
    }
    return 0;
}