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【高精度顺便可以水矩阵】【细菌】

2013-10-27 20:57:21| 分类： AC题目 | 标签： |举报 |字号大中小订阅

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Y博士发现了一种细菌，并命名为YYP细菌，他发现这种细菌有如下特性。
1、该细菌幼年期为3天，第四天便进化为成熟期细菌。
2、细菌进入成熟期，每天可产一个幼年期细菌。
在不考虑细菌寿命的情况下，第一天将1个细菌放入培养基中，请帮助Y博士计算一下第n天能够获得多少个细菌。

这一题可以直接得出递推式 F[N]=F[N-1]+F[N-3],初始状态为 F[1]=1,F[2]=1，F[3]=1,然后高精度加法即可。

program xijun;


type arr=array[0..100]of longint;
var f:array[1..150]of arr;
    n,i:longint;
function jia(a,b:arr):arr;
var i,j:longint;
begin
  j:=a[0];if j<b[0] then j:=b[0];
  for i:=1 to j+2 do jia[i]:=0;
  for i:=1 to j do
   begin
     jia[i]:=jia[i]+a[i]+b[i];
     jia[i+1]:=jia[i+1]+(jia[i] div 10);
     jia[i]:=jia[i] mod 10;
   end;
  inc(j);
  while (j>0)and(jia[j]=0) do dec(j);
  jia[0]:=j;
end;

begin
  read(n);
  for i:=1 to 3 do
   begin
   f[i][0]:=1;
   f[i][1]:=1;
   end;
  for i:=4 to n do
    f[i]:=jia(f[i-1],f[i-3]);
  for i:=f[n][0] downto 1 do
   write(f[n][i]);
end.

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