找出重复的起始点和重复周期,起始点不一定找第一个,可以使第一个周期中的任意一个。
#include<iostream> #include<stdio.h> using namespace std; int f[100000005]; int main() { int a,b,n,i,j; f[1]=1;f[2]=1; while(scanf("%d%d%d",&a,&b,&n)) { int s=0;//记录周期 if(a==0&&b==0&&n==0) break; for(i=3;i<=n;i++) { f[i]=(a*f[i-1]+b*f[i-2])%7; for(j=2;j<i;j++) if(f[i-1]==f[j-1]&&f[i]==f[j]) { s=i-j; //cout<<j<<" "<<s<<" >>"<<i<<endl; break; } if(s>0) break; } if(s>0){ f[n]=f[(n-j)%s+j]; //cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl; } cout<<f[n]<<endl; } return 0; }
因为f[i]只能取0~7,下面的程序用m[x][y],记录f[i]的值x y相邻时候出现过。
鸽巢原理知,状态总数不会超过7*7
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; const int N = 100; int f[N], m[8][8]; int main() { int n, a, b, k, x, y; while (scanf("%d%d%d", &a, &b, &n) != EOF && a+b+n) { memset(m, 0, sizeof(m)); f[1] = f[2] = x = y = 1; k = 3; while (!m[x][y]) { m[x][y] = k; cout<<"m["<<x<<"]"<<"["<<y<<"]="<<k<<endl; f[k] = y = (a * y + b * x) % 7; x = f[k-1]; k++; } int h = m[x][y];cout<<h<<" "<<k-h<<endl; if (n < k) printf("%d\n", f[n]); else printf("%d\n", f[(n-h)%(k-h)+h]); } }
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61632 Accepted Submission(s): 14087
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