45 lines
1.2 KiB
C++
45 lines
1.2 KiB
C++
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/*
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代码的作用是十进制的n转换成k进制的数字,输出的ans为进位的次数,len为结果的长度
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https://blog.csdn.net/qq_23109971/article/details/111396755
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*/
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 10010;
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int n, ans, k, len = 1; //最少是1位长度
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int d[N];
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int main() {
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cin >> n >> k; // n:要转换的10进制数,k:k进制
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//从0开始枚举每小于n的数字,一个个叠加上去
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for (int i = 0; i < n; i++) {
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++d[0]; //将叠加上来的1先放到个位上
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for (int j = 0; j < len - 1; j++) //看看会不会对现在的每一个位置造成进位?
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if (d[j] == k) {
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d[j] = 0;
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d[j + 1] += 1;
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++ans; //记录进位次数
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}
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//最高位单独处理
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if (d[len - 1] == k) {
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d[len - 1] = 0;
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d[len] = 1;
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++len;
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++ans;
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}
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//输出k进制的表示法
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// for (int i = len - 1; i >= 0; i--) cout << d[i];
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// cout << endl;
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}
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//输出进位次数
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// cout << ans << endl;
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cout << len << endl;
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//输出k进制的表示法
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// for (int i = len - 1; i >= 0; i--) cout << d[i];
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// cout << endl;
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return 0;
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}
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