Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 63888 Accepted Submission(s): 14701 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
我自己做的时候 忽略了n的大小,所以递归的时候栈溢出了.....
我的代码:
1 #include <iostream>
2 using namespace std;
3 static int count;
4 int f(int a,int b,int n);
5 int mod(int n);
6 int main()
7 {
8 int a,b,n;
9 while(cin>>a>>b>>n)
10 {
11 if(a>=1 && a<=1000 && b>=1 && b<=1000)
12 {
13 count=0;
14 cout<<f(a,b,n)<<endl;
15 }
16 else
17 break;
18 }
19 return 0;
20 }
21 int f(int a,int b,int n)
22 {
23 if( n == 1 )
24 return 1;
25 else if( n == 2)
26 return 1;
27 else
28 return mod(a*f(a,b,n-1)+b*f(a,b,n-2));
29 }
30 int mod(int n)
31 {
32 while(n>7)
33 {
34 n=n-7;
35 }
36 return n;
37 }
网上大神的代码:
#include <iostream>
#include <vector>
using namespace std;
vector<int> ivec;
//布尔数组元素flag[i][j]如果为true,则说明前面已经出现了i 和 j两者的组合
bool flag[7][7];
void init(void)
{
int i,j;
for(i = 0;i < 7;++i)
for(j = 0;j< 7;++j)
flag[i][j] = false;
}
int main()
{
int a,b,n;
while(cin>>a>>b>>n)
{
if(a == 0&&b == 0&&n == 0)
break;
ivec.clear();
init();
ivec.push_back(1);
ivec.push_back(1);
flag[1][1] = true;
int count = 1,f;
while(1)
{
f = (a*ivec.at(count)%7 + b*ivec.at(count - 1)%7)%7;
ivec.push_back(f);
++count;
//如果flag变量为true,则说明前面已经出现了这两者的组合,出现重复,无需下一步计算,直接break退出即可
if(flag[ivec.at(count)][ivec.at(count - 1)] == true)
break;
else
flag[ivec.at(count)][ivec.at(count - 1)] = true;
}
//count中存放的是ivec中出现循环前的元素总个数,注意ivec中的下标是从0开始计数的
count = count - 1;
if(n < count)
cout<<ivec.at(n-1)<<endl;
else
{
int j;
//for循环的目的是找出从那个地方开始重复,此处应该是从j处开始循环,注意j是从0下标开始计数的
for(j = 0;;++j)
if(ivec.at(count) == ivec.at(j) && ivec.at(count + 1) == ivec.at(j+1))
break;
n = (n - j)%(count - j);
if(n == 0)
n = count - j;
n += j;
cout<<ivec.at(n-1)<<endl;
}
}
return 0;
}