Bullseye

Problem

Maria has been hired by the Ghastly Chemicals Junkies (GCJ) company to help them manufacture bullseyes. A bullseye consists of a number of concentric rings (rings that are centered at the same point), and it usually represents an archery target. GCJ is interested in manufacturing black-and-white bullseyes. 

 

Maria starts with t millilitres of black paint, which she will use to draw rings of thickness 1cm (one centimetre). A ring of thickness 1cm is the space between two concentric circles whose radii differ by 1cm.

Maria draws the first black ring around a white circle of radius r cm. Then she repeats the following process for as long as she has enough paint to do so:

1. Maria imagines a white ring of thickness 1cm around the last black ring.
2. Then she draws a new black ring of thickness 1cm around that white ring.

Note that each "white ring" is simply the space between two black rings.

The area of a disk with radius 1cm is π cm². One millilitre of paint is required to cover area π cm². What is the maximum number of black rings that Maria can draw? Please note that:

• Maria only draws complete rings. If the remaining paint is not enough to draw a complete black ring, she stops painting immediately.
• There will always be enough paint to draw at least one black ring.

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of a line containing two space separated integers: r and t.

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the maximum number of black rings that Maria can draw.

Limits

Small dataset

1 ≤ T ≤ 1000.
1 ≤ r, t ≤ 1000.

Large dataset

1 ≤ T ≤ 6000.
1 ≤ r ≤ 10¹⁸.
1 ≤ t ≤ 2 × 10¹⁸.

Sample

Input 
  5
1 9
1 10
3 40
1 1000000000000000000
10000000000000000 1000000000000000000


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Output 
 Case #1: 1
Case #2: 2
Case #3: 3
Case #4: 707106780
Case #5: 49


https://code.google.com/codejam/contest/2418487/dashboard#s=p0

#include<stdio.h>#include<string.h>#include<stdlib.h>int main(){    long long ans,T,i,j,ncas=1,r,t,start,res;    FILE *file;    freopen("C:\\Users\\think\\Desktop\\in.in","r",stdin);    file=freopen("C:\\Users\\think\\Desktop\\output.txt","w",stdout);    scanf("%lld",&T);    while(T--)    {        scanf("%lld%lld",&r,&t);        start=ans=res=0;        start=2*r+1;        for(i=1;;i++)        {            ans=2*i*r+2*(i+1)*i+i+start;            if(ans>t)break;        }        //printf("%lld\n",i);        fprintf(file,"Case #%lld: %lld\n",ncas++,i);    }    return 0;}