TOJ 1203: Number Sequence
1203: Number Sequence

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
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 ;
}
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