主题链接:

题意:

就是给了一个公式,然后求出第n项是多少。。。

思路:

题目中n的范围实在是太大,所以肯定直接递推肯定会超时,所以想到的是暴力打表,找循环节,可是也不是那么easy发现啊,所以这时候分析一下,由于最后都会mod7,所以总共同拥有7X7总情况,即A 0,1,2,3,4,5,6,7。B也是如此,所以循环节为49,这么这个问题就攻克了。

。。

题目:

Number Sequence

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

Total Submission(s): 105674    Accepted Submission(s): 25691

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
 
Author
CHEN, Shunbao
 
Source
 
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代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std; #define mod 7 const int maxn=1000+10;
int a[maxn];
int A,B,n; int main()
{
while(~scanf("%d%d%d",&A,&B,&n))
{
if(A==0&&B==0&&n==0) return 0;
a[1]=1,a[2]=1;
for(int i=3;i<=100;i++)
a[i]=(A*a[i-1]+B*a[i-2])%mod;
printf("%d\n",a[n%49]);
}
return 0;
}

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