zoj1828||joj1949 解題報告

 

Fibonacci Numbers

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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.

高精度加法計算的問題，主要有一個問題就是空間複雜度，時間複雜度過得去，但是這道題第一次提交時，我卻超了內存，後來我只是加了三句delete語句就A了，我這才發現原來delete語句是這麼重要（以前都沒用過），僅僅是加了三句delete就使內存不超了。。。下面是A了的代碼，少一句delete都A不了！！！

代碼：

語言：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;//第一條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;//第二條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;//第三條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; }