Number Sequence 【打表】+【找規律】
阿新 • • 發佈:2019-02-20
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63052 Accepted Submission(s): 14475
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
Author CHEN, Shunbao
Source
Recommend JGShining
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
題意: 輸入 A,B,n 求f(n) 思路: 1 可以用陣列來模擬這種遞迴(居然才知道0.0) 2 一看 n 值那麼大,就知道 肯定會超時間按照一般方法 ,所以一定可以另闢蹊徑,所以就想到了,可能會有周期性即迴圈 所以就打表 一看 確實有規律,不過,不同的a,b,週期不同,所以 要打表來記錄 週期 3如果使用陣列或者遞迴都會爆棧,注意到題目中的數都是對7取餘,所以相鄰的兩個數的組合最多有49中,所以最多49次迴圈就會從頭開始迴圈,這是就可以不用繼續執行了,然後將n對i取餘,根據餘數就可以得到結果。
- #include<iostream>
- #include<stdio.h>
- usingnamespace 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;//記錄週期
-
if
- 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])// 由上述的4得
- {
- 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;
- }