87 lines
2.1 KiB
C++
87 lines
2.1 KiB
C++
#include<bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int a[N], cnt ;
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//求反数：数位分离数字，然后想办法再级成大数字
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bool HWS_1(int x) {
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//注意：反复使用不重新初始化肯定要出问题！！！！！
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//初始化
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memset(a, 0, sizeof a);
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cnt = 0;
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int t = x;
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//数位分离到一个数组中
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while (t) {
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int m = t % 10;
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a[cnt++] = m;
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t /= 10;
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}
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//方法（1）：通过数学公式POW拼成一个大数字
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int ans = 0;
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for (int i = 0; i < cnt; i++) ans += a[i] * pow(10, cnt - i - 1);//分析一下为什么是cnt-i-1
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return ans == x;
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}
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bool HWS_2(int x) {
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//注意：反复使用不重新初始化肯定要出问题！！！！！
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//初始化
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memset(a, 0, sizeof a);
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cnt = 0;
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int t = x;
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//数位分离到一个数组中
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while (t) {
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int m = t % 10;
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a[cnt++] = m;
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t /= 10;
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}
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//方法（2)：利用频繁的累乘方式拼成一个大数字
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int ans = 0;
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for (int i = 0; i < cnt; i++) ans = ans * 10 + a[i];
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return ans == x;
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}
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bool HWS_3(int x) {
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//注意：反复使用不重新初始化肯定要出问题！！！！！
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//初始化
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memset(a, 0, sizeof a);
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cnt = 0;
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int t = x;
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//数位分离到一个数组中
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while (t) {
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int m = t % 10;
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a[cnt++] = m;
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t /= 10;
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}
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//方法（3）:利用字符串转换
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string s = "";
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for (int i = 0; i < cnt; i++) s += to_string(a[i]);
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return to_string(x) == s;
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}
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//利用原始字符串反转进行
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bool HWS_4(int x) {
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string a = to_string(x);
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reverse(a.begin(), a.end());
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return a == to_string(x);
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}
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int main() {
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cout << HWS_1(12345) << endl;
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cout << HWS_2(12345) << endl;
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cout << HWS_3(12345) << endl;
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cout << HWS_4(12345) << endl;
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cout << endl;
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cout << HWS_1(12321) << endl;
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cout << HWS_2(12321) << endl;
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cout << HWS_3(12321) << endl;
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cout << HWS_4(12321) << endl;
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return 0;
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}
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