HDU 3944 DP? (Lucas定理)

题意：在杨辉三角中让你从最上面到 第 n 行，第 m 列所经过的元素之和最小，只能斜向下或者直向下走。

析：很容易知道，如果 m 在n的左半部分，那么就先从 (n, m)向左上，再直着向上，如果是在右半部分，那么就是先直着向上，再斜着左上。这样对应到，

左半部分：C(n, m) + C(n-1, m-1) + C(n-2, m-2) + ... + C(n-m, 0) + (n-m)

右半部分：C(n, m) + C(n-1, m) + C(n-2, m) + ... + C(m, m) + m

然后化简得到的答案就是C(n+1, m) + n - m，和C(n+1, m+1) + m。然后用Lucsa 定理就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <ctime>
#include <cstdlib>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL p;

LL quick_pow(LL a, LL n){
    LL ans = 1LL;
    a %= p;
    while(n){
        if(n & 1)  ans = ans * a % p;
        a = a * a % p;
        n >>= 1;
    }
    return ans;
}

LL C(LL n, LL m){
    if(n < m)  return 0;
    LL a = 1LL, b = 1LL;
    while(m){
        a = a * n % p;
        b = b * m % p;
        --n;  --m;
    }
    return a * quick_pow(b, p-2) % p;
}

LL Lucas(LL n, LL m){
    if(!m)  return 1;
    return C(n%p, m%p) * Lucas(n/p, m/p) % p;
}

int main(){
    int kase = 0;
    LL m, n;
    while(scanf("%I64d %I64d %I64d", &n, &m, &p) == 3){
        printf("Case #%d: ", ++kase);
        if(m * 2 < n)  printf("%I64d\n", (Lucas(n+1, m) + n - m) % p);
        else  printf("%I64d\n", (Lucas(n+1, m+1) + m) % p);
    }
}