华为题库
61.放苹果
#include <bits/stdc++.h>
using namespace std;
int process(int m, int n){
if(m < 0 || n < 0){
return 0;
}
else if(m == 1 || n == 1){
return 1;
}
else {
// 其他情况的时候，我们要计算少一个盘子，和所有盘子拿走一个苹果的情况
return process(m, n - 1) + process(m - n, n); //
}
}
int main(){
int m = 0, n = 0;
while(cin >> m >> n){
int res = process(m, n);
cout << res << endl;
}
return 0;
}
62.查找输入整数二进制中1的个数
#include<bits/stdc++.h>
using namespace std;
void process(int num, int& res){
while(num != 0){
res++;
num = num & (num - 1);
}
}
int main(){
int num = 0;
while(cin>>num){
int res = 0;
process(num, res);
cout<<res<<endl;
}
return 0;
}
63.DNA序列
#include <bits/stdc++.h>
using namespace std;
void process(string str, int num, string& res){
vector<string> vec; //保存所有可能的字串
for(int i = 0; i < str.size() - num + 1; i++){
string tmp = str.substr(i, num);
//cout << tmp << endl;
vec.push_back(tmp);
}
int maxLen = INT_MIN;
for(string s : vec){
int t = 0;
for(char ch : s){
if(ch == 'C' || ch == 'G'){
t++; //当前字串中C和G的个数
}
}
if(t > maxLen){
res = s;
maxLen = t;
}
}
}
int main(){
string str = "";
cin >> str;
int n = 0;
cin >> n;
string res = "";
process(str, n, res);
cout << res << endl;
return 0;
}
64.MP3光标位置
#include <bits/stdc++.h>
using namespace std;
int main(){
//输入
int musicNum = 0;
string command = "";
cin >> musicNum;
cin >> command;
//处理命令
if(musicNum <= 4){
int curMusicPos = 1; //当前光标的一个位置
for(char ch : command){
if(ch == 'U') curMusicPos--;
else curMusicPos++;
}
//输出
for(int i = 1; i <= musicNum; i++){
cout << i << " ";
}
cout << endl;
// 这里有一点要注意的就是有可能最后的位置超过了musicNum，要取一个模，变到这个范围内
if(curMusicPos >= 0){
cout << (curMusicPos) % musicNum << endl; //
}
else{
cout << (musicNum + curMusicPos + 1) % musicNum << endl;
}
}
else{
//musicNum > 4
int curMusicPos = 1; //当前光标的一个位置
int head = 1, end = 4; //head当前列表的头部是哪一首歌曲, end当前列表尾部是哪一首歌曲
for(char ch : command){
if(ch == 'U') {
curMusicPos--;
//如果光标比第一首歌曲的位置都要小
if(curMusicPos < 1){
//跳到最后一页的位置上面
curMusicPos = musicNum; //更新光标位置
head = musicNum - 4 + 1; //更新列表头部
end = musicNum; //更新列表尾部
}
else{
//如果光标没有小于1
if(curMusicPos < head){
//如果当前位置要比当前的列表头还要小，说明要更新列表了
head = curMusicPos; //更新列表头部
end--; //更新列表尾部
}
}
}
// ch == 'D'
else{
curMusicPos++;
//如果光标比最后一首歌曲的位置都要大
if(curMusicPos > musicNum){
//跳到第一页的位置上面
curMusicPos = 1; //更新光标位置
head = 1; //更新列表头部
end = 4; //更新列表尾部
}
else{
//如果光标没有大于musicNum
if(curMusicPos > end){
//如果当前位置要比当前的列表尾还要大，说明要更新列表了
end = curMusicPos; //更新列表尾部
head++; //更新列表头部
}
}
}
}
//输出
for(int i = head; i <= end; i++){
cout << i << " ";
}
cout << endl;
cout << curMusicPos << endl;
}
return 0;
}
65.查找两个字符串a,b中的最长公共子串
#include <bits/stdc++.h>
using namespace std;
//dp
//dp[i][j]表示 到s1第i个,到s2第j个为止 的公共子串长度 (其中s1较短)
void longestCommonSubsequence(string s1, string s2, string& res) {
vector<vector<int>> dp(s1.size() + 1, vector<int>(s2.size() + 1, 0));
int maxLength = INT_MIN, end = 0; //end表示字符串的末位位置 (最大不超过s1的长度)
for(int i = 1; i <= s1.size(); i++){
for(int j = 1; j <= s2.size(); j++){
if(s1[i - 1] == s2[j - 1]){
dp[i][j] = dp[i - 1][j - 1] + 1; //则增加长度
}
else{
dp[i][j] = 0; // //该位置为0
}
if(dp[i][j] > maxLength){
//更新最大长度
maxLength = dp[i][j];
end = i - 1; //
}
}
}
res = s1.substr(end - maxLength + 1, maxLength);
}
int main(){
string s1 = "";
string s2 = "";
getline(cin, s1);
getline(cin, s2);
if(s1.length() > s2.length()){
swap(s1, s2); //使较小的字符串在前
}
string res = "";
longestCommonSubsequence(s1, s2, res);
cout << res << endl;
return 0;
}
66.配置文件恢复
#include <bits/stdc++.h>
using namespace std;
vector<pair<string, string>> youXiaoCommand = {
{
"reset", ""},
{
"reset", "board"},
{
"board", "add"},
{
"board", "delete"},
{
"reboot", "backplane"},
{
"backplane", "abort"}
};
vector<string> commandRes = {
"reset what",
"board fault",
"where to add",
"no board at all",
"impossible",
"install first"
};
int main(){
string str = "";
while(getline(cin, str)){
stringstream iss(str);
string key1 = "", key2 = "";
getline(iss, key1, ' '); //第一个关键字
getline(iss, key2, ' '); //第二个关键字
string res = "";
int piPeiCount = 0; //记录匹配的关键字个数 (等于1时即唯一匹配到 即匹配成功)
for(auto iter = youXiaoCommand.begin(); iter != youXiaoCommand.end(); iter++){
int key1Pos = iter->first.find(key1); //key1在命令的前半部分第一次出现的下标 (为0匹配成功)
int key2Pos; //key2在命令的前半部分第一次出现的下标 (为0匹配成功)
if(key2 != ""){
//有key2
key2Pos = iter->second.find(key2); //key2在命令的前半部分第一次出现的下标
}
else if(key2 == "" && iter->second.empty()){
//特殊情况匹配到命令 {"reset", ""},
key2Pos = 0;
}
else{
//没匹配到
key2Pos = -1;
}
if(key1Pos == 0 && key2Pos == 0){
//匹配成功
piPeiCount++; //匹配成功的个数
res = commandRes[iter - youXiaoCommand.begin()]; //索引位置：iter - youXiaoCommand.begin()
}
}
if(piPeiCount == 1){
//匹配成功且为唯一匹配
cout << res << endl;
}
else{
//否则就是没有匹配成功，输出"unknown command"
cout<<"unknown command"<<endl;
}
}
return 0;
}
67.24点游戏算法
#include <bits/stdc++.h>
using namespace std;
double num = 1e-6;
int ANS = 24;
int ADD = 0, SUB = 1, MUL = 2, DIV = 3;
bool process(vector<double>& l){
if(l.size() == 0) return false;
if(l.size() == 1) return abs(24 - l[0]) < num; //只剩一个数字
int size = l.size(); //
for(int i = 0; i < size; i++){
for(int j = 0; j < size; j++){
if(i != j){
//两个索引不能相等
vector<double> l2;
for(int k = 0; k < size; k++){
if(k != i && k != j){
l2.push_back(l[k]);//先将本轮不做处理的数字放进去
}
}
//"二合一"
for(int m = 0; m < 4; m++){
//四种处理：加减乘除
if(m == ADD){
l2.push_back(l[i] + l[j]);
}
else if(m == SUB){
l2.push_back(l[i] - l[j]);
}
else if(m == MUL){
l2.push_back(l[i] * l[j]);
}
else{
if(abs(l[j]) < num) continue; //除的时候分母不能为0
l2.push_back(l[i] / l[j]);
}
if(process(l2)){
return true; //
}
l2.pop_back(); //
}
}
}
}
return false;
}
int main(){
int a = 0, b = 0, c = 0, d = 0;
while(cin >> a >> b >> c >> d){
//二合一 + dfs
vector<double> l1;
l1.push_back((double)a);
l1.push_back((double)b);
l1.push_back((double)c);
l1.push_back((double)d);
bool res = process(l1);
if(res)
cout << "true" << endl;
else
cout << "false" << endl;
}
return 0;
}
68.成绩排序
#include <bits/stdc++.h>
using namespace std;
struct cmpJiangXu{
bool operator()(const pair<string, int>& a, const pair<string, int>& b){
return a.second > b.second;
}
};
struct cmpShengXu{
bool operator()(const pair<string, int>& a, const pair<string, int>& b){
return a.second < b.second;
}
};
int main(){
int n = 0;
//法一
/*while(cin >> n){
int sortWay = 0; //排序方式
cin >> sortWay;
//输入名字和成绩
vector<pair<string, int>> info;
for(int i = 0; i < n; i++){
string name = "";
int score = 0;
cin >> name >> score;
info.push_back(make_pair(name, score));
}
//排序
//stable的函数可保证相等元素的原本相对次序在排序后保持不变
if(sortWay == 0){
//从高到低 降序
stable_sort(info.begin(), info.end(), cmpJiangXu());
}
else if(sortWay == 1){
//从低到高 升序
stable_sort(info.begin(), info.end(), cmpShengXu());
}
//输出
for(auto iter = info.begin(); iter != info.end(); iter++){
cout << iter->first << " " << iter->second << endl;
}
}*/
//法二
while(cin >> n){
int sortWay = 0; //排序方式
cin >> sortWay;
//输入名字和成绩
map<int, vector<string>> info;
for(int i = 0; i < n; i++){
//建立成绩和名字的映射
string name = "";
int score = 0;
cin >> name >> score;
info[score].push_back(name);
}
//排序
if(sortWay == 0){
//从高到低 降序
//逆序遍历 // rbegin rend
//输出
for(auto iter = info.rbegin(); iter != info.rend(); iter++){
//iter++
for(int i = 0; i < iter->second.size(); i++){
//成绩相同时，按照输入顺序输出
cout << iter->second[i] << " " << iter->first << endl;
}
}
}
else if(sortWay == 1){
//从低到高 升序
//顺序遍历
for(auto iter = info.begin(); iter != info.end(); iter++){
//iter++
for(int i = 0; i < iter->second.size(); i++){
cout << iter->second[i] << " " << iter->first << endl;
}
}
}
}
return 0;
}
69.矩阵乘法
#include <bits/stdc++.h>
using namespace std;
int main(){
//输入
int firstRows = 0;
int firstCols = 0;
int secondRows = 0;
int secondCols = 0;
cin >> firstRows;
cin >> firstCols;
secondRows = firstCols;
cin >> secondCols;
int num = 0;//矩阵元素
//输入第一个矩阵
vector<vector<int>> firstMatrix(firstRows, vector<int>(firstCols, 0));
for(int i = 0; i < firstRows; i++){
for(int j = 0; j < firstCols; j++){
cin >> num;
firstMatrix[i][j] = num;
}
}
//输入第二个矩阵
vector<vector<int>> secondMatrix(secondRows, vector<int>(secondCols, 0));
for(int i = 0; i < secondRows; i++){
for(int j = 0; j < secondCols; j++){
cin >> num;
secondMatrix[i][j] = num;
}
}
//计算相乘结果矩阵
vector<vector<int>> mulRes(firstRows, vector<int>(secondCols, 0));
for(int i = 0; i < firstRows; i++){
for(int j = 0; j < secondCols; j++){
for(int k = 0; k < firstCols; k++){
mulRes[i][j] += firstMatrix[i][k] * secondMatrix[k][j]; //相乘相加
}
}
}
//输出
for(int i = 0; i < firstRows; i++){
for(int j = 0; j < secondCols; j++){
cout << mulRes[i][j] << " ";
}
cout << endl; //换行，输出矩阵的下一行元素
}
return 0;
}
70.矩阵乘法计算量估算
#include <bits/stdc++.h>
using namespace std;
int main(){
int n = 0;
while(cin >> n){
vector<pair<int, int>> matrixs(n); //行数和列数
for(int i = 0; i < n; i++){
cin >> matrixs[i].first;
cin >> matrixs[i].second;
}
string s = "";
cin >> s;
//矩阵乘法 第一行第一列相乘相加
//A是一个50×10的矩阵，B是10×20的矩阵 则次数为：50×10×20
int res = 0;
stack<pair<int, int>> st; //栈存储矩阵的行数和列数
for(int i = 0; i < s.size(); i++){
//遍历字符串
if(s[i] == ')'){
//如果是右括号，则栈弹出两个元素，并累加乘法次数
auto y = st.top(); //后面的矩阵
st.pop();
auto x = st.top(); //前面的矩阵
st.pop();
if(x.second == y.first){
res += x.first * x.second * y.second;
st.push({
x.first, y.second}); // //把形成的新矩阵的行数和列数入栈
}
//
/*else if(y.second == x.first){
res += y.first * y.second * x.second;
st.push({y.first, x.second}); //把形成的新矩阵的行数和列数入栈
}*/
}
else if(s[i] != '('){
如果是字符，则直接入栈
//else if(isalpha(s[i])){
// A是第一个矩阵 它的行数和列数对应matrix[0]
int t = s[i] - 'A';
st.push(make_pair(matrixs[t].first, matrixs[t].second));
}
}
//输出
cout << res << endl;
}
return 0;
}