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sincerit 2117 Just a Numble(模擬除法)

2117 Just a Numble
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3817 Accepted Submission(s): 1915

Problem Description
Now give you two integers n m, you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed

Input
Each line of input will contain a pair of integers for n and m(1<=n<=107,1<=m<=105)

Output
For each line of input, your program should print a numble on a line,according to the above rules

Sample Input
4 2
5 7
123 123

Sample Output
5
0
8

題意求1/n 小數點後的第m位數字

#include <iostream>
#include <stdio.h> #include <cstring> #include <algorithm> using namespace std; #define N 100000 int main() { int n, m; while (cin >> n >> m) { if (n == 1) { cout << "0\n"; continue; } // 模擬除法 int t = 1; // 餘數 int k; for (int i = 1
; i <= m; i++) { t *= 10; // 餘數小於除數n 被除數÷除數=商 k = t / n; //小數點後第i位的商 t %= n; } cout << k << "\n"; } return 0; }