（hdu1005）Number Sequence（组合数学）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 176298 Accepted Submission(s): 43607

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 3

1 2 10

0 0 0

Sample Output

2

5

题意：定义：f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. 给A，B,n三个数，问f(n)是多少？

分析：这应该属于组合数学的问题，如果你注意到mod7，每个数的范围都是0-6，两个数组合一共有49种结果，然后直接递归

#include<cstdio>
#include<cstring>
using namespace std;
int f(int a,int b,int n)
{
if(n==1||n==2)
return 1;
else
return (a*f(a,b,n-1)+b*f(a,b,n-2))%7;
}
int main()
{
int n,a,b;
while(~scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)
{
printf("%d\n",f(a,b,n%49));
}
return 0;
}

优化：

记忆存储

#include<cstdio>
#include<cstring>
using namespace std;
const int N=50;
int A
;
int dfs(int a,int b,int n)
{
if(n==1||n==2) return 1;
int& ret=A
;
if(ret!=-1) return ret;
return ret=(a*dfs(a,b,n-1)+b*dfs(a,b,n-2))%7;
}
int main()
{
int n,a,b;
while(~scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)
{
memset(A,-1,sizeof(A));
printf("%d\n",dfs(a,b,n%49));
}
return 0;
}