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

 

1 # include
        
          2 # define inf 1000000007 3 const double eps=1e-8; 4 __int64 a[100005]; 5 void init(){ 6     int i; 7     for(i=0;i<=313;i++) 8         a[i] = 1; 9     for(i=314;i<=100000;i++){10         a[i]= a[i-100] + a[i-314];11         a[i]=a[i]%inf;12     }13 }14 int main(){15     int T,y,i;16     double x;17     init();18     scanf("%d",&T);19     while(T--){20         scanf("%lf",&x);21         if(x<0)22             printf("0\n");23         else if(x<1)24             printf("1\n");25         else26         {27             x=x*100;28             y=(int)(x+eps);29                 printf("%I64d\n",a[y]);30         }31     }32     return 0;33 }