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Time Limit:1000MS Memory Limit:32768KB
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
题解：
?因为n过大，f(n-1)和f(n-2)只有0,1,2,3,4,5,6七种情况，又A和B不变，f(n)由f(n-1)和f(n-2)决定，所以至多计算49次后，必将发生循环，所以计算50次即可。
以下是代码：
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#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <string>int c[1000];using namespace std;int main(){ int a,b,n; int t1,t2,t; c[1]=c[2]=1; while(scanf("%d%d%d",&a,&b,&n)!=EOF && (a || b || n)){ t1 =t2=1; int tag=0,i; for(i=3;i<=100;i++){ t = t2; t2 = (a*t + b*t1)%7; t1 = t ; c[i]=t2; if(t1==1 && t2==1)break;//发生循环，就终止。 } c[0]=c[i-2];//将循环的最后一个值赋值给c[0]，避免取余数后判断 printf("%d\n",c[n%(i-2)]);//i-2即周期 }} |
Winter-1-F Number Sequence 解题报告及测试数据
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原文地址：http://www.cnblogs.com/gzdaijie/p/4298092.html
