TOJ 1203: Number Sequence

1203: Number Sequence

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

1 1 3
1 2 10
0 0 0

Sample Output

2
5

Source

HDOJ

这个题可以用暴力打表，但是你不一定得到正确结果，然后啊，就挖出了一组破数据，7 7 n，这个第三项就是0了，然后都是0啊。就一直爆炸，RE、TLE各种姿势都不行的

所以循环的正确开始应该是f[3]、f[4]

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