2024-12-21【M-SOLUTIONS Programming Contest 2024(AtCoder Beginner Contest 232)】【题解A-D】

typedef long long ll;

typedef long double ld;

typedef pair<int,int> pii;

const int maxn=1e6+5;

void solve(){

string s,t;

cin >> s;

cin >> t;

int cur;

for(int i=0; i<s.size(); i++){

int del;

if(t[i] > s[i])del = t[i] - s[i];

else{

del = ‘z’ - s[i] + t[i] - ‘a’ + 1;

}

if(i == 0)cur = del;

else if(cur != del){

cout<<“No\n”; return;

}

}

cout<<“Yes\n”;

}

int main(){

ios::sync_with_stdio(0);

int t;

t=1 ;

while(t–){

solve();

}

return 0;

}

C - Graph Isomorphism

========================================================================================

题目描述

---

code:

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

typedef pair<int, int> pii;

const int maxn = 1e6 + 5;

void solve()

{

int n, m;

cin >> n >> m;

vector<unordered_set> mp(n);

for (int i = 0; i < m; i++)

{

int a, b;

cin >> a >> b;

a–, b–;

mp[a].insert(b);

mp[b].insert(a);

}

vector v(n);

for (int i = 0; i < n; i++)

v[i] = i;

vector edge(m);

for (int i = 0; i < m; i++)

{

cin >> edge[i].first >> edge[i].second;

}

do

{

bool flag = true;

for (int i = 0; flag && i < m; i++)

{

int a = v[edge[i].first - 1];

int b = v[edge[i].second - 1];

flag = mp[a].count(b);

}

if (flag)

{

cout << “Yes”;

return;

}

} whi

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le (next_permutation(v.begin(), v.end()));

cout << “No”;

}

int main()

{

ios::sync_with_stdio(0);

int t;

t=1;

while (t–)

{

solve();

}

return 0;

}

D - Weak Takahashi

=====================================================================================

题目描述

---

code:

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

typedef pair<int,int> pii;

const int maxn=1e6+5;

char mp[110][110];

bool rch[110][110];

void solve()

{

int h,w;

cin >> h >> w;

for(int i = 0;i < h;i++){

cin >> mp[i];

}

if(mp[0][0]==‘.’) rch[0][0] = true;

for(int i = 1;i < w;i++) if(mp[0][i]== ‘.’ && rch[0][i-1]) rch[0][i] = true;

for(int i = 1;i < h;i++) if(mp[i][0]== ‘.’ && rch[i-1][0]) rch[i][0] = true;

for(int i = 1;i < h;i++){

for(int j = 1;j < w;j++){

if(mp[i][j] == ‘.’ && (rch[i-1][j]||rch[i][j-1])) rch[i][j] = true;

}

}

int ans = 0;

for(int i = 0;i < h;i++){

for(int j = 0;j < w;j++){

if(rch[i][j]) ans = max(ans,i+j+1);

总结

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for(int j = 0;j < w;j++){

if(rch[i][j]) ans = max(ans,i+j+1);

总结

在清楚了各个大厂的面试重点之后，就能很好的提高你刷题以及面试准备的效率，接下来小编也为大家准备了最新的互联网大厂资料。

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