Problem Description
There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)
Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.
Sample Input
1
2
3
Sample Output
1
2
4
#include <iostream>
#include <string>
std::string add(std::string strA, std::string strB);
int main(int argc, const char *argv[])
{
std::string strResult[1000] = {"1", "2", "4", "7"};
for(int i = 4;i < 1000;++ i)
{
strResult[i] = add(add(strResult[i - 1], strResult[i - 2]), strResult[i - 4]);
}
int n = 0;
while(std::cin >> n)
{
std::cout << strResult[n - 1] << std::endl;
}
//system("pause");
return 0;
}
std::string add(std::string strA, std::string strB)
{
int nLenMax, nLenMin;
std::string strMax, strMin;
if(strA.size() >= strB.size())
{
nLenMax = strA.size();
strMax = strA;
nLenMin = strB.size();
strMin = strB;
}
else
{
nLenMax = strB.size();
strMax = strB;
nLenMin = strA.size();
strMin = strA;
}
for(int i = nLenMin - 1, j = nLenMax - 1;i >= 0;-- i, -- j)
{
strMax[j] += strMin[i] - '0';
}
for(int i = nLenMax - 1;i > 0;-- i)
{
if(strMax[i] > '9')
{
strMax[i] -= 10;
++ strMax[i - 1];
}
}
if(strMax[0] > '9')
{
strMax[0] -= 10;
strMax = '1' + strMax;
}
return strMax;
}