union Data{ int num; float value; char symbol; }; union Data data;
struct Node { int data; Node* next; };
现在有一个指向链表头部的指针:Node* head。如果想要在链表中插入一个新节点,其成员data的值为42,并使新节点成为链表的第一个节点,下面哪个操作是正确的?( )
#include<iostream>
#include<cmath>
using namespace std;
double f(double a,double b,double c){
double s=(a+b+c)/2;
return sqrt(s*(s-a)*(s-b)*(s-c));
}
int main(){
cout.flags(ios::fixed);
cout.precision(4);
int a,b,c;
cin>>a>>b>>c;
cout<<f(a,b,c)<<endl;
return 0;
}
假设输入的所有数都为不超过1000的正整数,完成下面的判断题和单选题:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int f(string x,string y){
int m=x.size();
int n=y.size();
vector<vector<int>>v(m+1,vector<int>(n+1,0));
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(x[i-1]==y[j-1]){
v[i][j]=v[i-1][j-1]+1;
}else{
v[i][j]=max(v[i-1][j],v[i][j-1]);
}
}
}
return v[m][n];
}
bool g(string x,string y){
if(x.size() != y.size()){
return false;
}
return f(x+x,y)==y.size();
}
int main(){
string x,y;
cin>>x>>y;
cout<<g(x,y)<<endl;
return 0;
}
分析:
#include <iostream>
#include <cmath>
using namespace std;
int solve1(int n){
return n*n;
}
int solve2(int n){
int sum=0;
for(int i=1;i<=sqrt(n);i++){
if(n%i==0){
if(n/i==i){
sum+=i*i;
}else{
sum+=i*i+(n/i)*(n/i);
}
}
}
return sum;
}
int main(){
int n;
cin>>n;
cout<<solve2(solve1(n))<<" "<<solve1((solve2(n)))<<endl;
return 0;
}
假设输入的n是绝对值不超过1000的整数,完成下面的判断题和单选题。
#include <iostream>
#include <vector>
using namespace std;
int find_missing(vector<int>& nums){
int left = 0, right = nums.size() - 1;
while (left < right){
int mid = left + (right-left) / 2;
if (nums[mid]==mid+ ①){
②;
}else{
③
}
}
return ④;
}
int main(){
int n;
cin >> n;
vector<int> nums(n);
for (int i= 0; i< n; i++) cin >> nums[i];
int missing_number = find_missing(nums);
if(missing_number == ⑤) {
cout << "Sequence is consecutive" << endl;
}else{
cout << "Missing number is " << missing numbeer << endl;
}
return 0;
}
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int min(int x,int y,int z){
return min(min(x,y),z);
}
int edit_dist_dp(string str1,string str2){
int m=str1.length();
int n=str2.length();
vector<vector<int>> dp(m+1,vector<int>(n+1));
for(int i=0;i<=m;i++){
for(int j=0;j<=n;j++){
if(i==0)
dp[i][j]=①;
else if(j==0)
dp[i][j]=②;
else if(③)
dp[i][j]=④;
else
dp[i][j]=1+min(dp[i][j-1],dp[i-1][j],⑤);
}
}
return dp[m][n];
}
int main(){
string str1,str2;
cin>>str1>>str2;
cout<<"Mininum number of operation:"
<<edit_dist_dp(str1,str2)<<endl;
return 0;
}