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

代码如下:
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]);
}
}
}
}
}
}

我输入0 0 0时eclipse显示错误;不明白怎么改,求大神帮忙

3个回答

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;

huruiba
坎特 回复眼里一片海: 回复眼里一片海: 这个题,主要是数组下标是从0开始算的,而函数f从f1 f2这样开始,我贴代码给你,好好学,基础要扎实
2 年多之前 回复
m0_37632283
眼里一片海 十分感谢 ; 但是最后在杭电运行显示错误答案。这是我的代码,还是不明白哪里出错了
2 年多之前 回复

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+3];
a[1] = 1;
a[2] = 1;
if(n==1||n==2) {
System.out.println(1);;
}
else if(A==0&&B==0&&n==0) {
continue;
}
else 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]);
}
}
}
}
}
}

public static void main(String[] args) {
Scanner 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];//保存函数值 这里应该是n+2,不是n+3
//数组下标从0开始算
a[0] = 1;//f(1)
a[1] = 1;//f(2)
if (n == 1 || n == 2) {
System.out.println(1);
} else if (A == 0 && B == 0 && n == 0) {
//continue;
break;
} else if (A > 1000 & n > 1000000000) {//这个条件是干什么用的
System.exit(0);
} else if (A >= 1 && B <= 1000 && n >= 3 && n <= 1000000000) {
for (int t = 2; t <= n; t++) {
//int m = t-1;
// f(n) = (A * f(n - 1) + B * f(n - 2)) % 7;
// f(3) = (a * f(2)+ b* f(1))*7
a[t] = ((A * a[t - 1]) + (B * a[t - 2])) % 7;
//System.out.println(a[t]);
if (t == (n - 1)) {//你的错误主要是在这里,t==n是不对的,t是数组的下标,从0开始,而函数f的值从1开始
System.out.println(a[t]);
}
}
}
}

}
m0_37632283
眼里一片海 在t==n时a[t]中的值存放已经是最终的f(n)的值了,这里并没有错误,我调试运行后跟杭电例子的输入输出是一样的。 上方如果a[n]设的是a[n+2]的大小话,那么当n==0的时候,下方a[2]就没有值可以存入了.......
2 年多之前 回复
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