1. 程式人生 > >TOJ 1203: Number Sequence

TOJ 1203: Number Sequence

contains http tom log .html clu image ons medium

1203: Number Sequence 技術分享 技術分享

Time Limit(Common/Java):1000MS/10000MS Memory Limit:65536KByte
Total Submit: 957 Accepted:208

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

Source

HDOJ

這個題可以用暴力打表,但是你不一定得到正確結果,然後啊,就挖出了一組破數據,7 7 n,這個第三項就是0了,然後都是0啊。就一直爆炸,RE、TLE各種姿勢都不行的

所以循環的正確開始應該是f[3]、f[4]

#include <stdio.h>
int main(){
int a,b,n,i;
int f[55]={0,1,1};
while(scanf("%d%d%d",&a,&b,&n),a||b||n){
    f[3]=(a+b)%7
; f[4]=(a*f[3]+b)%7; for(i=5;i<=n;i++){ f[i]=(a*f[i-1]+b*f[i-2])%7; if(f[i]==f[4]&&f[i-1]==f[3]){ n=(n-3)%(i-4)+3; break; } } printf("%d\n",f[n]); } return 0; }

TOJ 1203: Number Sequence