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题目地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1828
Time Limit: 2 Seconds Memory Limit: 65536 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 Input100
Sample Output354224848179261915075
Note:No generated Fibonacci number in excess of 1000 digits will be in the test data, i.e. f(20) = 6765 has 4 digits.
考察高精度运算
关键点是求每一位相加后的数字,并且还应该保证在输出的时候尽可能的方便。
这个时候关键点就是对a[i][j]进行对10取余,这样就保证了数组正序存放的数是逆序的数据。
这道题竟然是多组输入,找了 很久BUG,发现是输出错误~~~~~~
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdio>
using namespace std;
int a[5010][1010];
void init()
{
a[1][0]=1;
a[2][0]=1;
for(int i=3; i<=5001; ++i)
{
int c=0;
for(int j=0;j<=1001;++j)
{
a[i][j]=(a[i-1][j]+a[i-2][j]+c)%10;
c=(a[i-1][j]+a[i-2][j]+c)/10;
}
}
}
int main()
{
init();
int n;
while(~scanf("%d",&n))
{
int i;
for(i=1001;i>=0;--i)
if(a[n][i]!=0)
break;
for(;i>=0;--i)
{
printf("%d",a[n][i]);
}
printf("\n");
}
return 0;
}