HDOJ 1250 Hat's Fibonacci(大数，错了11次！！)
发布时间：2020-12-14 02:26:43 所属栏目：大数据 来源：网络整理
导读：Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9103????Accepted Submission(s): 2953 Problem Description A Fibonacci sequence is calculated by adding the previous two
Hat's FibonacciTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
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.
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Output
For each case,output the result in a line.
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Sample Input
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Sample Output
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神题，各种超RT，各种运行超限。。。 ? 题目没有直接给出n的最大值，只给出了数列中最大数长度不超过2005，据大牛测试n为7060时，长度为2012。这样就知道数组长度 至少为7060。然而并没有什么乱用，常规做法还是会超限。 这里可以用数组的每一个纵向下标储存4位数解决超限问题。 ? 上代码： ? #include<cstdio>
#define num 8000
#define max 510
int a[num][max];
void fib()
{
int i,j,c,s;
a[1][max-1]=a[2][max-1]=a[3][max-1]=a[4][max-1]=1;
for(i=5;i<num;i++)
{
c=0;
for(j=max-1;j>=0;j--)
{
s=a[i][j]+a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]+c;
a[i][j]=s%10000;
c=s/10000;
}
}
}
int main()
{
fib();
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<max;i++)
if(a[n][i]!=0)
break;
printf("%d",a[n][i]);
i++;
for(;i<max;i++)
printf("%04d",a[n][i]);
printf("n");
}
return 0;
}
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