Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 115124 Accepted Submission(s): 27959
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).
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
Recommend
这道题真的让我觉得有些头大，题意简单明了，就是按照它的公式推出后来的项目，然后提交发现是超时的，这个时候才发现后面的那个n非常大，但是通项，通项又求不出来，只好但是看的mod 7 觉得应该可以找到规律了吧，嗯，然后就找，那如果两个 1 再次出现不久刚好是个循环吗，但到这里还是把问题想的太简单了，提交了一次RE掉了，后来差到题解，才发现还这是不够的如果出现 7 7 n 这种例子，后面将全是0，然后数组不管开多大都会爆炸，所以不应该用题目已经给你了的值，应该根据输入的情况定出检查的两个值。所以修改后应该是推后两位去取值
#include<cstdio>
#define maxn 60
using namespace std;
int f[maxn] = {0,1,1};
int main(){
int a,b,c;
while(~scanf("%d %d %d",&a,&b,&c) && a && b && c){
int cnt;
for(int i = 3;i < 54;i++){
f[i] = (f[i-1]*a + f[i-2]*b)%7;
if(i > 5){ //取值定在5之上是为了防止前面的重复区段
if(f[i-1] == f[3] && f[i] == f[4]){
cnt = i - 4;
break;
}
}
}
if(c > 2){
printf("%d\n",f[(c-3)%cnt + 3]); //这段代码是为了防止取值取到零
}else{
printf("1\n");
}
}
return 0;
}
