【HDU 3483】 A Very Simple Problem (二项式展开+矩阵加速)

A Very Simple Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1022    Accepted Submission(s): 500

Problem Description

This is a very simple problem. Given three integers N, x, and M, your task is to calculate out the following value:

Input

There are several test cases. For each case, there is a line with three integers N, x, and M, where 1 ≤ N, M ≤ 2*10⁹, and 1 ≤ x ≤ 50.
The input ends up with three negative numbers, which should not be processed as a case.

Output

For each test case, print a line with an integer indicating the result.

Sample Input

100 1 10000

3 4 1000

-1 -1 -1

Sample Output

5050

444

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

【分析】

　　感觉我的思想离正解还是太远太远，根本不会这样想。

　　首先我们发现x很小，k很大，k达到了10^9，for一遍都不行，这样我们就想到ans可能存在递推式，然后可以用矩阵加速。

　　

转自：http://972169909-qq-com.iteye.com/blog/1863402

　　太厉害了，我自己都好难再推一次~~

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define Maxn 60
#define LL long long

LL c[Maxn][Maxn];
LL n,x,M;

struct node
{
    LL a[Maxn][Maxn];
}t[5];

void init()
{
    memset(c,0,sizeof(c));
    for(LL i=0;i<=50;i++) c[i][0]=1;
    for(LL i=1;i<=50;i++)
     for(LL j=1;j<=50;j++)
         c[i][j]=(c[i-1][j-1]+c[i-1][j])%M;
}

void get_un()
{
    memset(t[1].a,0,sizeof(t[1].a));
    for(LL i=0;i<=x+1;i++) t[1].a[i][i]=1%M;
}

void mul(LL now,LL y,LL z)
{
    for(int i=0;i<=x+1;i++)
     for(int j=0;j<=x+1;j++)
     {
         t[2].a[i][j]=0;
         for(LL k=0;k<=x+1;k++)
          t[2].a[i][j]=(t[2].a[i][j]+t[y].a[i][k]*t[z].a[k][j])%M;
     }
    t[now]=t[2];
}

void qpow(LL b)
{
    get_un();
    while(b)
    {
        if(b&1) mul(1,0,1);
        mul(0,0,0);
        b>>=1;
    }
}

int main()
{
    while(1)
    {
        scanf("%lld%lld%lld",&n,&x,&M);
        if(n==-1&&x==-1&&M==-1) break;
        init();
        for(LL i=0;i<=x;i++)
        {
            for(LL j=0;j<=i;j++) t[0].a[i][j]=(x*c[i][j])%M;
            for(LL j=i+1;j<=x+1;j++) t[0].a[i][j]=0;
        }
        for(LL i=0;i<x;i++) t[0].a[x+1][i]=0;
        t[0].a[x+1][x]=t[0].a[x+1][x+1]=1%M;
        // get_un();
        qpow(n+1);
        printf("%lld
",t[1].a[x+1][0]);
    }
    return 0;
}

2016-09-25 22:07:05