hdu 1005——Number Sequence
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水题。但并不容易我是看题解才知道周期的,我也知道有周期可求不出来!!!!!
大神物语:凡是给出运算公式的数学题,只要没有优化的话,超时超内存是避免不了的!
如果直接根据式子写出for循环,则会爆栈。所以自己要根据给的式子找周期,该题明显是找规律的题,找它的运算周期;由关系式可知每一项只与它的前两项有关,
所以当连续的两项在前面出现过,则循环节出现。对于公式f[n]=a*f[n-1]+b*f[n-2];后者只有7*7=49种可能。
为什么这么说,因为对于f[n-1]或f[n-2]的取值只有0,1,2,3,4,5,6,7种可能
又因为a,b固定,所以前后相乘产生49种可能。
一组测试数据中最坏的情况是49种情况都遇到,那么会在第50次运算中产生循环节
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 115443 Accepted Submission(s): 28057
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 31 2 100 0 0
Sample Output
25
#include<stdio.h>int main(){ int a,b,n,m,t; int f[1010]={0,1,1}; while(scanf("%d%d%d",&a,&b,&n),a,b,n) { for(int i=3;i<49;i++) f[i]=(a*f[i-1]+b*f[i-2])%7; printf("%d\n",f[n%49]); } return 0;}
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