杭電1005 http://acm.hdu.edu.cn/showPRoblem.php?pid=1005 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
算法講解： 1、由于f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7，則(A * f(n - 1)) mod 7有7種可能， (B * f(n - 2)) mod 7有7種可能，所以f(n)有7*7=49種結(jié)果。 2、注意n=1及n=2時f(n)=1，以后計算49次，再以后就重復前面的結(jié)果了。 vc2010測試代碼如下：
#include<stdio.h>void main(){ long n; int A,B,ret[52]; while(1) { scanf("%d%d%ld",&A,&B,&n); if(A==0 && B==0 && n==0) break; if(n==1 || n==2) { printf("1/n"); } else { ret[1]=1; ret[2]=1; for(int i=3; i<52; i++) { ret[i]=(A*ret[i-1]+B*ret[i-2])%7; } for(int i=0; i<3; i++) { ret[i]=ret[49+i]; } printf("%d/n",ret[n%49]); } }}新聞熱點
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