Winter-1-F Number Sequence 解题报告及测试数据

Time Limit:1000MS Memory Limit:32768KB

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

题解：

?因为n过大，f(n-1)和f(n-2)只有0,1,2,3,4,5,6七种情况，又A和B不变，f(n)由f(n-1)和f(n-2)决定，所以至多计算49次后，必将发生循环，所以计算50次即可。

以下是代码：

#include <iostream>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <string>

int c[1000];

using namespace std;

int main(){

int a,b,n;

int t1,t2,t;

c[1]=c[2]=1;

while(scanf("%d%d%d",&a,&b,&n)!=EOF && (a || b || n)){

t1 =t2=1;

int tag=0,i;

for(i=3;i<=100;i++){

t = t2;

t2 = (a*t + b*t1)%7;

t1 = t ;

c[i]=t2;

if(t1==1 && t2==1)break;//发生循环，就终止。

}

c[0]=c[i-2];//将循环的最后一个值赋值给c[0]，避免取余数后判断

printf("%d\n",c[n%(i-2)]);//i-2即周期

}

时间： 2024-10-10 22:11:41

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