Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 104190 Accepted Submission(s): 25232
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
题目主要是要找循环节。
一开始找规律于是又了下面的代码
1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int a[100]; 7 int main() 8 { 9 //freopen("in.txt","r",stdin); 10 int n,A,B; 11 while(scanf("%d %d %d",&A,&B,&n)&&(n||A||B)) 12 { 13 a[1]=1; 14 a[2]=1; 15 for(int i=3;i<=49;i++) 16 a[i]=(A*a[i-1]+B*a[i-2])%7; 17 printf("%d ",a[n%49]); 18 } 19 return 0; 20 }
后来发现如果A=7，B=7，n>2的话，输入都应该会是0，但是实际输出为1。虽然数据有点水，所以还是要严谨对待。
做法还是找循环节，只是说多了很多细节，要好好体会。
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 const int MAXN=200+10; 8 int a[MAXN]; 9 int main() 10 { 11 //freopen("in.txt","r",stdin); 12 int A,B,n,i; 13 a[1]=a[2]=1; 14 while(scanf("%d %d %d",&A,&B,&n)!=EOF) 15 { 16 if(A==B&&B==n&&n==0)break; 17 int flag=0; 18 for(i=3;i<=200;i++) 19 { 20 a[i]=(A*a[i-1]+B*a[i-2])%7; 21 if(a[i]==1&&a[i-1]==1) 22 break; 23 if(a[i]==0&&a[i-1]==0){ 24 flag=1;break; 25 } 26 } 27 if(flag) 28 { 29 printf("0 "); 30 continue; 31 } 32 if(i>n) 33 { 34 printf("%d ",a[n]); 35 continue; 36 } 37 i=i-2; 38 n=n%i; 39 if(n==0)n=i; 40 printf("%d ",a[n]); 41 } 42 return 0; 43 }