HDU 1005.Number Sequence【很多问题是不能直接求的】【8月15】
2015-08-15 22:10
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Number Sequence
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
Sample Output
f[1]=f[2]=1;f
=(A*f[n-1]+B*f[n-2])%7;给定A,B,n,求f
。
不能直接求,不然肯定超时超内存。寻找循环。f[1]=1,f[2]=1;当连续出现两个1的时候就说明是循环了。代码如下:
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
f[1]=f[2]=1;f
=(A*f[n-1]+B*f[n-2])%7;给定A,B,n,求f
。
不能直接求,不然肯定超时超内存。寻找循环。f[1]=1,f[2]=1;当连续出现两个1的时候就说明是循环了。代码如下:
#include<cstdio> int main(){ int a,b,n,f[11000]; f[1]=1;f[2]=1; while(scanf("%d%d%d",&a,&b,&n)!=EOF,a||b||n){ int i; for(i=3;i<11000;i++){ f[i]=(a*f[i-1]+b*f[i-2])%7; if(f[i]==1&&f[i-1]==1)//i-2就是循环长度 break; } f[0]=f[i-2]; printf("%d\n",f[n%(i-2)]); } return 0; }
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