poj2696--A Mysterious Function（记忆化递归)

根据题目写出取模运算，跟着公式做记忆化递归

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 1 //Accepted    392K    188MS    G++    908B
 2 //Accepted    400K    172MS    G++
 3 #include <stdio.h>
 4 #include <math.h>
 5 #include <string.h>
 6 
 7 int a,b,c,d,e,f,g,h,i;
 8 int ans[1001];
 9 int mod(int a,int b)
10 {
11     int r;
12     for(r=0;r<=b-1;r++)
13         if((a-r)%b==0)
14             return r;
15 }
16 int solve(int n)//顺推过去 
17 {
18     for(int k = 3; k <= n; k++)
19     {
20             if(k % 2 == 1)
21             {
22                  int t = d * ans[k-1] + e * ans[k-2] - f * ans[k-3];
23                  ans[k] = mod(t,g); 
24             }
25             else
26             {
27                   int t =  f * ans[k-1] - d * ans[k-2]  + e * ans[k-3];
28                   ans[k] = mod(t,h);
29             }
30     }
31 }
32 int solve1(int n)//递归思路 
33 {
34     if(ans[n] > -1) return ans[n];
35     else if(n % 2 == 1)
36     {
37           int t = d * solve1(n-1) + e * solve1(n-2) - f * solve1(n-3);
38           ans[n] = mod(t,g); 
39           return ans[n];
40     }
41     else
42     {
43           int t =  f * solve1(n-1) - d * solve1(n-2)  + e * solve1(n-3);
44           ans[n] = mod(t,h);
45           return ans[n];
46     }
47 }
48 int main(void)
49 {
50     int t;
51     scanf("%d",&t);
52     while(t--)
53     {
54               memset(ans,-1,sizeof(ans));
55               scanf("%d%d%d%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f,&g,&h,&i);
56               ans[0] = a;
57               ans[1] = b;
58               ans[2] = c;
59               solve1(i);
60               printf("%d\n",ans[i]);
61     }
62     return 0;    
63 }
posted @ 2012-09-18 16:07  Wheat″  阅读(220)  评论(0)  编辑  收藏  举报