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.
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 }