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杭电ACM 1021 Fibonacci Again
Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14170 Accepted Submission(s): 6581
Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
Output
Print the word "yes" if 3 divide evenly into F(n).
Print the word "no" if not.
Print the word "no" if not.
Sample Input
0 1 2 3 4 5
Sample Output
no no yes no no no
#include <stdio.h> int main(){ int n, t; while (scanf("%d", &n) == 1){ t = n&7; printf("%s\n", t==2 || t==6 ? "yes" : "no"); } return 0; }
杭电ACM 1005 Number Sequence
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38553 Accepted Submission(s): 8150
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 3 1 2 10 0 0 0
Sample Output
2 5
#include <stdio.h> #include <string.h> int main(){ int b[7][7]; int a[100]; int n; int x, y, m, st, len; while (scanf("%d %d %d", &x, &y, &m) == 3){ if (x==0 && y==0 && m==0){ break; } a[1] = 1; a[2] = 1; memset(b, 0, sizeof(b)); b[1][1] = 1; n = 3; for (;;){ a[n] = (a[n-1]*x+a[n-2]*y)%7; if (b[a[n-1]][a[n]] != 0){ break; } b[a[n-1]][a[n]] = n-1; ++n; } st = b[a[n-1]][a[n]]; len = n-1-st; if (m < st){ printf("%d\n", a[m]); } else{ printf("%d\n", a[st+(m-st)%len]); } } return 0; }