HDU 1893 Sibonacci Numbers（斐波那契）

Sibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 706    Accepted Submission(s): 242

Problem Description

As is known to all, the definition of Fibonacci Numbers is: 
f(1)=1 
f(2)=1 
f(n)=f(n-1)+f(n-2) (n>=3) 

Now Sempr found another Numbers, he named it "Sibonacci Numbers", the definition is below: 
f(x)=0 (x<0) 
f(x)=1 (0<=x<1) 
f(x)=f(x-1)+f(x-3.14) (x>=1) 

Your work is to tell me the result of f(x), is the answer is too large, divide it by 1000000007 and give me the remainder. 
Be careful the number x can be an integer or not. 

 

 

Input

In the first line there is an Integer T(0<T<=10000) which means the number of test cases in the input file. 
Then followed T different lines, each contains a number x(-1000<x<1000). 

 

 

Output

For each case of the input file, just output the result, one for each line. 

 

 

Sample Input

3

-1

0.667

3.15

 

 

Sample Output

0

1

2

 

巧妙点在于以0.01为下标，哈哈，这是相当不理解的吧

再有就是精度问题，错了好几次的

 

# include<stdio.h>
# define inf 1000000007
const double eps=1e-8;
__int64 a[100005];
void init(){
    int i;
    for(i=0;i<=313;i++)
        a[i] = 1;
    for(i=314;i<=100000;i++){
        a[i]= a[i-100] + a[i-314];
        a[i]=a[i]%inf;
    }
}
int main(){
    int T,y,i;
    double x;
    init();
    scanf("%d",&T);
    while(T--){
        scanf("%lf",&x);
        if(x<0)
            printf("0\n");
        else if(x<1)
            printf("1\n");
        else
        {
            x=x*100;
            y=(int)(x+eps);
                printf("%I64d\n",a[y]);
        }
    }
    return 0;
}