NYOJ427Number Sequence
Number Sequence
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
- 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).
- 输入
- 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.
- 输出
- For each test case, print the value of f(n) on a single line.
- 样例输入
-
1 1 3 1 2 10 0 0 0
- 样例输出
-
2 5
View Code1 2 #include<stdio.h> 3 int main() 4 { 5 int f[100000]; 6 int a,b,n,i; 7 while(scanf("%d%d%d",&a,&b,&n)&&(a||b||n)) 8 { 9 10 f[1]=1;f[2]=1; 11 if(n==1||n==2){printf("1\n");continue;} 12 for(i=3;i<100000;i++) 13 { 14 f[i]=(a*f[i-1]+b*f[i-2])%7; 15 if(f[i]==1&&f[i-1]==1) break; 16 } 17 i=i-2; 18 f[0]=f[i]; 19 printf("%d\n",f[n%i]); 20 } 21 return 0; 22 } 23
posted on 2012-11-14 21:42 LinuxPanda 阅读(270) 评论(0) 编辑 收藏 举报