十进制的分数转化为七进制的小数

F - Consecutive Digits

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit  Status  Practice  POJ 2710

Description

As a recruiting ploy, Google once posted billboards in Harvard Square and in the Silicon Valley area just stating “{first 10-digit prime found in consecutive digits of e}.com”. In other words, find that 10-digit sequence and then connect to the web site — and find out that Google is trying to hire people who can solve a particular kind of problem. 

Not to be outdone, Gaggle (a loosy-goosy fuzzy logic search firm), has devised its own recruiting problem. Consider the base 7 expansion of a rational number. For example, the first few digits of the base 7 expansion of 1/5  ₁₀ = 0.12541...  ₇, 33/4  ₁₀ = 11.15151...  ₇, and 6/49  ₁₀ = 0.06000...  ₇, From this expansion, find the digits in a particular range of positions to the right of the "decimal" point. 

Input

The input file begins with a line containing a single integer specifying the number of problem sets in the file. Each problem set is specified by four base 10 numbers on a single line, n d b e, where n and d are the numerator and denominator of the rational number and 0 <= n <= 5,000 and 1 <= d <= 5,000. b and e are the beginning and ending positions for the desired range of digits, with 0 <= b,e <= 250 and 0 <= (e-b) <= 20. Note that 0 is the position immediately to the right of the decimal point.

Output

Each problem set will be numbered (beginning at one) and will generate a single line: 

• Problem set k: n / d, base 7 digits b through e: result

where k is replaced by the problem set number, result is your computed result, and the other values are the corresponding input values. 

Sample Input

4
1 5 0 0
6 49 1 3
33 4 2 7
511 977 122 126

Sample Output

Problem set 1: 1 / 5, base 7 digits 0 through 0: 1
Problem set 2: 6 / 49, base 7 digits 1 through 3: 600
Problem set 3: 33 / 4, base 7 digits 2 through 7: 151515

Problem set 4: 511 / 977, base 7 digits 122 through 126: 12425

#include   #include   #include   #include #include #include #define MAX 255 #include #include using namespace std;      int main() { int T,n,d,b,e; cin>>T; for(int i=1;i<=T;i++) { cin>>n>>d>>b>>e; printf("Problem set %d: %d / %d, base 7 digits %d through %d: ",i,n,d,b,e); n%=d; for(int j=0;j<=e;j++) { n*=7; if(j>=b) printf("%d",n/d); n%=d; } cout< } return 0; }