sincerit 2117 Just a Numble(模擬除法)
阿新 • • 發佈:2018-11-05
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;
}