Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 159015 Accepted Submission(s): 39002
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
题解:f[n]每次都%7,那么这样就会有规律。所以先打表找到周期
注意:若A=B=0,则n>=3,f[n]=0;所以要进行判定;
找到周期的标志就是从i=3开始,若还有f[i]=f[i-1]=1,就开始了之前的循环。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int main()
{
int A,B,n;
int f[220];
while(~scanf("%d%d%d",&A,&B,&n))
{
bool flag=true;
if(A==0&&B==0&&n==0)
break;
f[1]=f[2]=1;
int i;
for(i=3;i<=200;i++) //打表找到周期
{
f[i]=(A*f[i-1]+B*f[i-2])%7;
if(f[i]==1&&f[i-1]==1) //找到周期退出
break;
if(f[i]==0&&f[i-1]==0) //A=B=0,则f[...]=0
{
flag=false;
break;
}
}
if(i>n) // i 至少是 3,所以这一个判断应该在判断flag之前
{
printf("%d\n",f[n]);
continue;
}
if(!flag)
{
printf("0\n");
continue;
}
i-=2; //周期等于 i-2,减去的是后来相等的 i和i-1
n=n%i;
if(n==0)
n=i;
printf("%d\n",f[n]);
}
return 0;
}
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