解题报告：HDU_3944 DP? 数论

DP?

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

Total Submission(s): 2833    Accepted Submission(s): 880


[align=left]Problem Description[/align]



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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

1 1 2
4 2 7

 

[align=left]Sample Output[/align]

Case #1: 0
Case #2: 5

 

[align=left]Author[/align]
phyxnj@UESTC
 

[align=left]Source[/align]
2011 Multi-University Training Contest 11 - Host by
UESTC
 

[align=left]Recommend[/align]
xubiao
 

Statistic | Submit | Discuss
|
Note

题意：在杨辉三角上，给出n,k,p。问以0,0为起点到n,k处的最小和模p的值。

分析：

step 1.

先由组合的对称性可知k与n-k的结果一样，所以这里我们只考虑k>n/2的情况：当k>n/2时，最优的线路是先斜走到k,k ，然后竖直向下走到n,k，根据组合公式

①


②


得出当k>n/2时，最优线路的总和值为(C(n+1,k+1)+k)%p，然后就转换成组合数求模问题。

step 2.

如果p是固定值，那么相信大家都会做了：打p内的阶乘表及阶乘逆元表，然后用lucas很快就能查询出结果。
虽然p不是固定的，但是打素数表可以发现1e4的范围内只有1250个素数，那么我们可以打出这些素数全部表，然后根据p的值用对应的表就行了（题目给的内存特别大）。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

namespace prime_table{

const int MAX_N = 1e4;
const int MAX_M = sqrt(MAX_N+0.5);

int all=0;
int pr[MAX_N/10+300];
bool isp[MAX_N+10];

inline void init(){
memset(isp,0,sizeof(isp));
for(int i=2;i<=MAX_N;i++){
if(!isp[i]){pr[all++]=i;
if(i<=MAX_M)
for(int j=i*i;j<=MAX_N;j+=i){
isp[j]=true;
}
}
}
//    printf("all=%d\n",all);
}

}

using namespace prime_table;

namespace Combinatorial{
const int N = 10005;/** n 的最大 值 **/
int fac[N+5][1300],ifac[N+5][1300];

/**快速幂**/
int q_pow(int a,int b,int mod){
int n=1;
while(b){
if(b&1){
n=(int)(1LL*n*a%mod);
}
a=(int)(1LL*a*a%mod);
b>>=1;
}
return n;
}
/**预处理逆元**/
inline void init_QP(){
for(int i=0;i<all;i++){
fac[0][i]=ifac[0][i]=1;
for(int j=1;j<=pr[i];j++){
fac[j][i]=(int)(1LL*fac[j-1][i]*j%pr[i]);
ifac[j][i]=q_pow(fac[j][i],pr[i]-2,pr[i]);
}
}
}

int C(int a,int b,int id){
if(b>a || b<0)return 0;
return (int)(1LL*fac[a][id]*ifac[b][id]%pr[id]*ifac[a-b][id]%pr[id]);
}

int lucas(int a,int b,int id){
if(!a||!b){
return 1;
}
return (int)(1LL*lucas(a/pr[id],b/pr[id],id)*C(a%pr[id],b%pr[id],id)%pr[id]);

}
}

using namespace Combinatorial;

int main(){
init();init_QP();
int n,k,p,t=0;
while(scanf("%d%d%d",&n,&k,&p)==3){
int *id = lower_bound(pr,pr+all,p);
k = max(k,n-k);
printf("Case #%d: %d\n",++t,(lucas(n+1,k+1,id-pr)+k)%p);

}return 0;
}