网易互娱模拟笔试交流

贴一下我的答案，写的不整洁，全是暴力求解😂。

第一题求交税，暴力就行了：

#include<iostream>
using namespace std;

int main() {
int T; cin >> T;
while (T--) {
int earning; cin >> earning;
double leve11 = 0, level2 = 0, level3 = 0, level4 = 0, level5 = 0, level6 = 0, level7 = 0;
if (earning > 5000) {
leve11 = earning > 8000 ? (3000 * 0.03) : (earning - 5000)*0.03;
}
if (earning > 8000) {
level2 = earning > 17000 ? (9000 * 0.10) : (earning - 8000)*0.10;
}
if (earning > 17000) {
level3 = earning > 30000 ? (13000 * 0.20) : (earning - 17000)*0.20;
}
if (earning > 30000) {
level4 = earning > 40000 ? (10000 * 0.25) : (earning - 30000)*0.25;
}
if (earning > 40000) {
level5 = earning > 60000 ? (20000 * 0.30) : (earning - 40000)*0.30;
}
if (earning > 60000) {
level6 = earning > 85000 ? (25000 * 0.35) : (earning - 60000)*0.35;
}
if (earning > 85000) {
level7 = (earning - 85000)*0.45;
}
double res = leve11 + level2 + level3 + level4 + level5 + level6 + level7;
if (res - (int)res >= 0.5) {
cout << res + 1 << endl;
}
else {
cout << res << endl;
}
}
return 0;
}

第二题字符串缩写，如ABCD缩写A-D，DABCDEFGX缩写DA-GX，暴力就行了，但是我感觉我写的很乱：

#include<iostream>
#include<string>
using namespace std;

int main() {
int T; cin >> T;
while (T--) {
string s; cin >> s;
if (s.size() < 4) {
cout << s << endl;
}
else {
string res = "";
int i = 0;
int j = i;
for (; j < s.size() - 1;) {
if (s[j + 1] - s[j] == 1) {
j++;
}
else {
if (j - i + 1 >= 4) {
res += s[i]; res += '-'; res += s[j];
}
else {
for (int k = i; k <= j; k++)
res += s[k];
}
j++; i = j;
}
}
if (j - i + 1 >= 4) {
res += s[i]; res += '-'; res += s[j];
}
else {
res += s.substr(i);
}

cout << res << endl;
}
}
return 0;
}

第三题：把一个正整数N的X进制和Y进制表示连在了一起输出,得到了一个无法识别的数字，请还原这个数，如13 7 1016，原来的数为13，输入T组数据

用内置pow函数出错，不知道为啥，换了乘法就行了，还是暴力求解：

#include<iostream>
#include<string>
using namespace std;

int main() {
int T; cin >> T;
while (T--) {
int X, Y; string Z;
cin >> X >> Y; cin >> Z;

for (int i = 0; i < Z.size() - 1; i++) {
string s1 = Z.substr(0, i + 1);
string s2 = Z.substr(i + 1);

int num1 = 0, num2 = 0;
for (int i = 0; i <= s1.size() - 1; i++) {
num1 *= X;
if (s1[i] >= 'A'&& s1[i] <= 'F')
num1 += (s1[i] - 'A' + 10);
else
num1 += (s1[i] - '0');
}
for (int i = 0; i <= s2.size() - 1; i++) {
num2 *= Y;
if (s2[i] >= 'A'&& s2[i] <= 'F')
num2 += (s2[i] - 'A' + 10);
else
num2 += (s2[i] - '0');
}

if (num1 == num2) {
cout << num1 << endl;
break;
}
}
}
return 0;
}