HDUOJ Number Sequence 题目1005


/*Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 127146 Accepted Submission(s): 30901

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).

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

Author

CHEN, Shunbao

Source

ZJCPC2004

Recommend

JGShining | We have carefully selected several similar problems for you: 1008 1001 1003 1009 1012

*/

#include<stdio.h>

int main()

{

int a,b,i;

long long f[55],n;

while(1)

{ scanf("%d %d %lld",&a,&b,&n);

if(a==0&&b==0&&n==0)break;

f[1]=f[2]=1;

for(i=3;i<=49;i++)

f[i]=(a*f[i-1]+b*f[i-2])%7;

printf("%d\n",f[n%48]);

}

return 0;

}/*Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 127146 Accepted Submission(s): 30901

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).

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

Author

CHEN, Shunbao

Source

ZJCPC2004

Recommend

JGShining | We have carefully selected several similar problems for you: 1008 1001 1003 1009 1012

*/

又一道找规律题目

#include<stdio.h>

int main()

{

int a,b,i;

long long f[55],n;

while(1)

{ scanf("%d %d %lld",&a,&b,&n);

if(a==0&&b==0&&n==0)break;

f[1]=f[2]=1;

for(i=3;i<=49;i++)

f[i]=(a*f[i-1]+b*f[i-2])%7;

printf("%d\n",f[n%48]);

}

return 0;

}