TBACAMILLE 2018-01-20 12:43 采纳率: 50%

# 杭电OJ1005 不明白错误在哪

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 187350 Accepted Submission(s): 46661

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

import java.util.*;

public class Main{
private static Scanner in;

public static void main(String[] args) {

in = new Scanner(System.in);
while(in.hasNext())
{
int A = in.nextInt();
int B = in.nextInt();
int n = in.nextInt();
int []a = new int[n+2];
a[1] = 1;
a[2] = 1;
if(n==1||n==2) {
System.out.println(1);;
}
if(A==0&&B==0&&n==0) {
continue;
}
if(A1000&n>1000000000) {
System.exit(0);
}
else if(A>=1&&B<=1000&&n>=3&&n<=1000000000)
{
for(int t = 3;t<=n;t++)
{
a[t] = ((A*a[t-1])+(B*a[t-2]))%7;
if(t==n) {
System.out.println(a[t]);
}
}
}
}
}
}

• 写回答

#### 3条回答默认 最新

• 坎特 2018-01-20 12:50
关注

把对0 0 0的判断拉到前面
if(A==0&&B==0&&n==0) {
continue;
}
放在循环体开头
错在数组下标越界吧，n=0的时候，数组只有a[0]，a[1]才对
int []a = new int[n+2];
a[1] = 1;
a[2] = 1;

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