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Hat's Fibonacci(大数,好)

时间:2015-09-12 18:53:30      阅读:217      评论:0      收藏:0      [点我收藏+]

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Hat‘s Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9394    Accepted Submission(s): 3065


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
题解:大数相加超内存,所以想着每个里面存8位;输出的时候不足8位补0;
代码:
 1 #include<stdio.h>
 2 #include<string.h>
 3 const int MAXN=10100;
 4 const int MAXM=100000000;
 5 char temp[MAXN];
 6 int dp[MAXN][510];
 7 int main(){
 8     dp[0][0]=1;
 9     dp[1][0]=1;
10     dp[2][0]=1;
11     dp[3][0]=1;
12     dp[4][0]=1;
13     for(int i=5;i<=10000;i++){
14         for(int j=0;j<=500;j++)dp[i][j]=dp[i-1][j]+dp[i-2][j]+dp[i-3][j]+dp[i-4][j];
15         for(int j=0;j<=500;j++)
16             if(dp[i][j]>MAXM)dp[i][j+1]+=dp[i][j]/MAXM,dp[i][j]%=MAXM;    
17     }
18     int n;
19     while(~scanf("%d",&n)){
20         int j=500;
21         while(!dp[n][j])j--;
22         printf("%d",dp[n][j]);
23         while(--j>=0)printf("%08d",dp[n][j]);//不足8位补零 
24         puts("");
25     }
26     return 0;
27 }

 

Hat's Fibonacci(大数,好)

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原文地址:http://www.cnblogs.com/handsomecui/p/4803355.html

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