大斐波那契数杭电1250
Hat's Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9518 Accepted Submission(s): 3135
Problem Description
A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.
Input
Each line will contain an integers. Process to end of file.
Output
For each case, output the result in a line.
Sample Input
100
Sample Output
4203968145672990846840663646
Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
#include<stdio.h>
#include<string.h>
using namespace std;
struct big
{
int a[500];
int t;
}f[7050];
//设置数组简直口怕.
int main()
{
int i,j,tmp;//tmp用来表示余数和进位数.
for(i=0;i<7050;i++)
{
f[i].t=1;
}
f[1].a[0]=f[2].a[0]=f[3].a[0]=f[4].a[0]=1;
/////////////初始化完毕.
for(i=5;i<7050;i++)
{
f[i].t=f[i-1].t;
for(tmp=j=0;j<f[i].t;j++)
{
f[i].a[j]=(f[i-1].a[j]+f[i-2].a[j]+f[i-3].a[j]+f[i-4].a[j]+tmp)%1000000; //取6位数字
tmp=(f[i-1].a[j]+f[i-2].a[j]+f[i-3].a[j]+f[i-4].a[j]+tmp)/1000000; //记录进位
}
if(tmp!=0)
{
f[i].a[f[i].t]=tmp;
f[i].t++;
}
}
while (scanf("%d",&i) == 1)
{
j=f[i].t-1;
printf("%d",f[i].a[j]);//打印的时候防止输入1的时候输出000001
for(j--;j>=0;j--)
{
printf("%06d",f[i].a[j]); //打印时用0补齐位数
}
printf("\n");
}
return 0;
}