3-19 博乐科技C++笔试部分题解
全是暴力的题也没啥可说的。
就是第一题一直只过60%。给我恶心吐了
第四题手写个堆，话说过了就行谁要自己写堆呀。
第一题求税率
第二题求拦截水最大
第三题求字符串S能拆除几个T
第四题排序
第二题：
class Solution {
public:
int maxArea(vector<int>& height) {
// write code here
int ans=0;
int n = height.size();
for(int i =0;i<n;i++){
for(int j =i+1;j<n;j++){
ans = max(ans,min(height[i],height[j])*(j-i));
}
}
return ans;
}
};
第三题
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 输入s为一个长度为n的字符串，t是一个长度为m的字符串， 1<= s.length <= 10**9, 1 <= t.length <= 10**9。
* @param s string字符串
* @param t string字符串
* @return int整型
*/
int MatchString(string s, string t) {
// write code here
unordered_map<char, int>mpS;
unordered_map<char, int>mpT;
unordered_map<char, int>mp;
for(int i = 0;t[i];i++){
mpT[t[i]] ++;
}
for(int i =0;s[i];i++) mpS[s[i]]++;
int ans = 10000000;
for(int i = 0;t[i];i++){
mp[t[i]]++;
if(mp[t[i]] == 1){
ans = min(ans,mpS[t[i]] / mpT[t[i]]);
}
}
return ans;
}
};
4
class Solution {
public:
vector<int> heap_top_k(vector<int>& nums, int k) {
// write code here
vector<int>ans;
sort(nums.begin(),nums.end());
for(int i =nums.size()-1;i > nums.size() - k-1;i--) ans.push_back(nums[i]);
return ans;
}
}; 第一题恶心心，只能过60% ，我直接改了一个多小时，改上头了
double nowNum = 0.0;
class Solution {
public:
string toString(double num){
string ans = "";
num *= 100;
num += 0.5;
long long p = (long long)num;
long long x = p;
int cnt = 0;
while(p){
ans.push_back(p % 10 + '0');
cnt++;
if(cnt == 2) ans.push_back('.');
p /= 10;
}
if(x <= 9){
ans.push_back('0');
ans.push_back('.');
ans.push_back('0');
}
else if(x <= 99){
ans.push_back('0');
}
reverse(ans.begin(), ans.end());
return ans;
}
double getNum(double needCost){
double ans = 0.0;
if(nowNum + needCost > 960000.0){
double p = min(needCost,(nowNum + needCost - 960000));
ans += p * 0.45;
needCost -= p;
}
if(needCost < 0.01) return ans;
if(nowNum + needCost > 660000.0){
double p = min(needCost,(nowNum + needCost - 660000));
ans += p * 0.35;
needCost -= p;
}
if(needCost < 0.01) return ans;
if(nowNum + needCost > 420000.0){
double p = min(needCost,(nowNum + needCost - 420000));
ans += p * 0.30;
needCost -= p;
}
if(needCost < 0.01) return ans;
if(nowNum + needCost > 300000){
double p = min(needCost,(nowNum + needCost - 300000));
ans += p * 0.25;
needCost -= p;
}
if(needCost < 0.01) return ans;
if(nowNum + needCost > 144000.0){
double p = min( (nowNum + needCost - 144000),needCost );
ans += p * 0.20;
needCost -= p;
}
if(needCost <= 0.01) return ans;
if(nowNum + needCost > 36000){
double p = min ( needCost, (nowNum + needCost - 36000) );
ans += p * 0.1;
needCost -= p;
}
if(needCost <= 0.01) return ans;
if(nowNum + needCost > 0){
ans += ( needCost) * 0.03;
}
return ans;
}
vector<string> calcTax(vector<double>& income) {
nowNum = 0.0;
vector<string>ans;
for(int i=0; i < income.size(); i++){
if(income[i] <= 5000){
ans.push_back("0.00");
}
else {
double realNum = getNum( income[i] - 5000 );
string x = toString(realNum);
if(x == "" || realNum < 0.01) x = "0.00";
nowNum += (income[i] - 5000);
ans.push_back(x);
}
}
return ans;
}
}; 
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