zoj1828||joj1949 解题报告

Fibonacci Numbers

Time Limit: 1 Second Memory Limit: 32768 KB

A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1.

f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2)

Your task is to take a number as input, and print that Fibonacci number.


Sample Input

100

Sample Output

354224848179261915075

Note:

No generated Fibonacci number in excess of 1000 digits will be in the test data, i.e. f(20) = 6765 has 4 digits.

高精度加法计算的问题，主要有一个问题就是空间复杂度，时间复杂度过得去，但是这道题第一次提交时，我却超了内存，后来我只是加了三句delete语句就A了，我这才发现原来delete语句是这么重要（以前都没用过），仅仅是加了三句delete就使内存不超了。。。下面是A了的代码，少一句delete都A不了！！！

代码：

语言：c++

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char *m1,*m2,*m3;
long n;
m1=new char[1010*sizeof(char)];
m2=new char[1010*sizeof(char)];
char *addition(char *m1,char *m2);
while(cin>>n)
{
m1[0]='1';
m1[1]='/0';
m2[0]='1';
m2[1]='/0';
if(n==1||n==2)
cout<<1<<endl;
else
{
for(int i=3;i<=n;++i)
{
m3=addition(m1,m2);
strcpy(m1,m2);
strcpy(m2,m3);
delete m3;//第一条delete
}
cout<<m2<<endl;
}
}
return 0;
}
char *addition(char *m1,char *m2)
{
int *str2int(char *str);
int check(int *a,int n);
char *int2str(int *a,int n);
int len1=strlen(m1),len2=strlen(m2),len,i,*t1,*t2;
len=(len1>=len2)?len1:len2;
t1=str2int(m1);
t2=str2int(m2);
for(i=len1;i<=len;++i)
t1[i]=0;
for(i=len2;i<=len;++i)
t2[i]=0;
for(i=0;i<len;++i)
t1[i]+=t2[i];
delete t2;//第二条delete
len=check(t1,len);
return int2str(t1,len);
}
int *str2int(char*str)
{
int len=strlen(str);
int *a=new int[(len+1)*sizeof(int)];
for(int i=0;i<len;++i)
a[i]=int(str[len-1-i])-'0';
return a;
}
char *int2str(int *a,int n)
{
int len=n;
char *str=new char[(len+1)*sizeof(char)];
for(int i=0;i<len;++i)
str[i]=char(a[len-1-i])+'0';
str[len]='/0';
delete a;//第三条delete
return str;
}
int check(int *a,int n)
{
int len=n,i;
while(a[len-1]==0&&len>1)
--len;
for(i=0;i<len;++i)
{
if(a[i]>=10)
{
a[i+1]+=a[i]/10;
a[i]%=10;
}
}
if(a[i]!=0)
len=i+1;
return len;
}