*杭电1005——Number Sequence(数学题)
2015-10-22 17:12
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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
看到题目,一开始用的递归,因为n过大,肯定会发生栈溢出。后来尝试用数组从小到大一个一个计算,开一个100,000,000的数组超过了题目设置的内存大小。然后,百度了下,因为有mod7,所以f(n-1)和f(n-2)的值只有7种,即为0~6.f
的值在最多在f[49] 7*7=49之后会出现重复。(为什么最多49个元素后面出现循环,即第49~97个元素与第1~48个元素一一对应相等,为什么是49明白,但是为什么出现循环就不太清楚~~~)
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
看到题目,一开始用的递归,因为n过大,肯定会发生栈溢出。后来尝试用数组从小到大一个一个计算,开一个100,000,000的数组超过了题目设置的内存大小。然后,百度了下,因为有mod7,所以f(n-1)和f(n-2)的值只有7种,即为0~6.f
的值在最多在f[49] 7*7=49之后会出现重复。(为什么最多49个元素后面出现循环,即第49~97个元素与第1~48个元素一一对应相等,为什么是49明白,但是为什么出现循环就不太清楚~~~)
# include<stdio.h> int f[50]; int main() { int A,B,n; while(1) { scanf("%d%d%d",&A,&B,&n); if(A==0&&B==0&&n==0) break; else { int i; f[1]=1; f[2]=1; for(i=3;i<=49;i++) { f[i]=(A*f[i-1]+B*f[i-2])%7; if(f[i]==1&&f[i-1]==1) break;//进入下一次循环,i-2即为一次循环元素的个数(最大为49) } n=n%(i-2); if(n==0) printf("%d\n",f[i-2]); else printf("%d\n",f ); } } return 0; }
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