hdoj 2031 进制转换

来源：互联网发布：大华图像算法工程师编辑：程序博客网时间：2024/06/05 16:37

进制转换

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37621 Accepted Submission(s): 20634

Problem Description

输入一个十进制数N，将它转换成R进制数输出。

Input

输入数据包含多个测试实例，每个测试实例包含两个整数N(32位整数)和R（2<=R<=16, R<>10）。

Output

为每个测试实例输出转换后的数，每个输出占一行。如果R大于10，则对应的数字规则参考16进制（比如，10用A表示，等等）。

Sample Input

7 223 12-4 3

Sample Output

1111B-11

Author

lcy

Source

C语言程序设计练习（五）

转了个大圈，晕........

#include<iostream>#include<cstdio>#include<string>#include<sstream>#include<cstdlib>#include<cstring>#include<cctype>#include<algorithm>#include<stack>#include<queue>#include<map>#include<set>#include<vector>#include<deque>#include<cmath>#include<climits>#include<list>#include<utility>#include<memory>#include<cstddef>#include<iterator>using namespace std;typedef long long LL;const double pi = acos(-1.0);///////////////////////////////vector<int>vec;void solve(int num,int num2) {if (num / num2 == 0) {vector<int>::iterator a;a= vec.begin();vec.insert(a, num);}else {vector<int>::iterator a;a = vec.begin();vec.insert(a, num%num2);solve(num / num2, num2);}return;}///////////////////////////////int main(int argc, char**argv) {//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);////////////////////////////int num;while (cin >> num) {vec.clear();int num2;cin >> num2;int flag = 0;if (num < 0) {flag = 1;num = -num;}solve(num, num2);vector<int>::iterator a1,b1;a1 = vec.begin();b1 = vec.end();if (flag == 1) cout << "-";for (; a1 != b1; a1++) {if (*a1 >= 10) {int t = *a1;char temp = t - 10 + 'A';cout << temp;}else {cout << *a1;}}cout << endl;}//////////////////////////////system("pause");return 0;}

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