网易9.12 C++开发,最后两个大题,7的倍数和编辑距离
7的倍数:
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int N = 100010;
int V[N];
int f[N][7];
int main()
{
int n = 0;
while(cin>>V[n++],V[n-1]);
memset(f,-1,sizeof f);
f[0][0] = 0;
for(int i = 0;i < n;i ++){
for(int j = 0;j < 7;j ++){
f[i+1][j] = max(f[i+1][j],f[i][j]) ;
if(f[i][j] == -1) continue;
int t = (j + V[i])%7;
f[i+1][t] = max(f[i+1][t],f[i][j]+V[i]);
}
}
cout<<f[n][0];
return 0;
} 编辑距离: #include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
unordered_set<string> set;
vector<string> getarr(string& S){
int idx = 0;
vector<string> res;
while(idx < S.size()){
string arr;
while(idx < S.size()&&S[idx] != ' '){
arr += S[idx];
idx++;
}
idx++;
if(arr.size()) res.push_back(arr);
}
return res;
}
void dp(vector<string>& A,vector<string>& B){
int n = A.size(),m = B.size();
vector<vector<int>> f(n+1,vector<int>(m+1,1e9));
f[0][0] = 0;
for(int i = 1;i <= n;i ++)
if(set.count(A[i-1])) f[i][0] = f[i-1][0];
else f[i][0] = f[i-1][0]+1;
for(int i = 1;i <= m;i ++)
if(set.count(B[i-1])) f[0][i] = f[0][i-1];
else f[0][i] = f[0][i-1]+1;
for(int i = 1;i <= n;i ++){
for(int j = 1;j <= m;j ++){
if(set.count(A[i-1])){
f[i][j] = min(f[i][j],f[i-1][j]);
}
if(set.count(B[j-1])){
f[i][j] = min(f[i][j],f[i][j-1]);
}
if(set.count(A[i-1])&&set.count(B[j-1])){
f[i][j] = min(f[i][j],f[i-1][j-1]);
}
if(A[i-1] == B[j-1]){
f[i][j] = min(f[i-1][j-1],f[i][j]);
f[i][j] = min(f[i-1][j]+1,f[i][j]);
f[i][j] = min(f[i][j-1]+1,f[i][j]);
}
else{
f[i][j] = min(f[i-1][j-1]+1,f[i][j]);
f[i][j] = min(f[i-1][j]+1,f[i][j]);
f[i][j] = min(f[i][j-1]+1,f[i][j]);
}
}
}
cout<<f[n][m];
}
int main()
{
string temp1;
getline(cin,temp1);
int idx = 0;
while(idx < temp1.size()){
string arr;
while(idx < temp1.size()&&temp1[idx] != ' '){
arr += temp1[idx];
idx++;
}
idx++;
set.insert(arr);
}
string A,B;
getline(cin,A),getline(cin,B);
auto VA = getarr(A);
auto VB = getarr(B);
dp(VA,VB);
return 0;
}