Hat‘s Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7854 Accepted Submission(s): 2551
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#include<stdio.h> #include<string.h> #define N 10000 //N很关键,N=10000或者20000都可以AC,N取得过小会错。 int str[N][260]; int main() { memset(str,0,sizeof(str)); str[1][0]=1; str[2][0]=1; str[3][0]=1; str[4][0]=1; int i,j,ans=0,c,n; for(i=5;i<N;i++) { for(j=0,c=0;j<260;j++) { ans=str[i-1][j]+str[i-2][j]+str[i-3][j]+str[i-4][j]+c; c=ans/100000000; str[i][j]=ans%100000000; //每一个数组存8位数字,c来控制是否进位。 } } while(scanf("%d",&n)!=EOF) { j=259; while(!str[n][j]) //将首0清除。 j--; printf("%d",str[n][j]); for(i=j-1;i>=0;i--) printf("%08d",str[n][i]);//每8位输出,不足8位自动补0; printf("\n"); } return 0; }
Hat's Fibonacci(杭电1250)
时间: 2024-10-11 12:21:00