【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[];

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

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

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

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

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

[HDU 3483]

2016-09-25 22:07:05