AtCoder Beginner Contest 351
A
easy
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| #include<bits/stdc++.h>
using ll=long long;
using namespace std;
int main()
{
ll sum=0;
ll ans=0;
for(int i=1;i<=9;i++)
{
ll x;
cin>>x;
sum+=x;
}
for(int i=1;i<=8;i++)
{
ll x;
cin>>x;
ans+=x;
}
cout<<max(0ll,sum-ans+1);
return 0;
}
|
B
easy
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| #include<bits/stdc++.h>
using ll=long long;
using namespace std;
char a[102][102];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin>>a[i][j];
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
char x;
cin>>x;
if(x!=a[i][j])
{
cout<<i<<' '<<j;
return 0;
}
}
}
return 0;
}
|
C
用一个栈模拟
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| #include<bits/stdc++.h>
using ll=long long;
using namespace std;
stack<int>a;
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
int x;
cin>>x;
if(a.empty())
{
a.push(x);
continue;
}
if(x==a.top())
{
while(!a.empty()&&x==a.top())
{
a.pop();
x++;
}
a.push(x);
}
else
{
a.push(x);
}
}
cout<<a.size();
return 0;
}
|
D
dfs，不过有一个地方就是走过的点那他肯定不是最优的，就直接Vis掉。
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| #include<bits/stdc++.h>
#pragma GCC optimize("O1,O2,O3,O4,O5,Og,Os,Ofast,inline")
using namespace std;
int n,m;
struct point {
int x,y;
};
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
char ch[1005][1005],vis[1000005],vis2[1000005];
queue<point> q;
vector<int> x_,y_;
int bfs(int x,int y) {
point p = {x,y};
q.push(p);
vis[p.x*1000+p.y-1] = 1,x_.push_back(p.x),y_.push_back(p.y);
int ans = 0;
while(!q.empty()) {
point tmp = q.front();
q.pop();
ans++;
if(ch[tmp.x][tmp.y-1]=='#'||ch[tmp.x-1][tmp.y]=='#'||ch[tmp.x][tmp.y+1]=='#'||ch[tmp.x+1][tmp.y]=='#'||ch[tmp.x][tmp.y]=='#') continue;
vis2[tmp.x*1000+tmp.y-1] = 1;
for(int i = 0; i<4; i++) {
point np = tmp;
np.x+=dx[i];
np.y+=dy[i];
if(1<=np.x&&np.x<=n&&1<=np.y&&np.y<=m&&!vis[np.x*1000+np.y-1]&&!vis2[np.x*1000+np.y-1]) q.push(np),vis[np.x*1000+np.y-1] = 1,x_.push_back(np.x),y_.push_back(np.y);
}
}
for(int i = x_.size()-1;i>=0;i--) vis[x_[i]*1000+y_[i]-1] = 0;
x_.clear();
y_.clear();
return ans;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
int ans = 0;
for(int i = 1; i<=n; i++)
for(int j = 1; j<=m; j++)
cin >> ch[i][j],ans = ans||(ch[i][j]=='.');
for(int i = 1; i<=n; i++)
for(int j = 1; j<=m; j++)
if((!(ch[i][j-1]=='#'||ch[i-1][j]=='#'||ch[i][j+1]=='#'||ch[i+1][j]=='#'||ch[i][j]=='#'))&&!vis2[i*1000+j-1])
ans = max(ans,bfs(i,j));
cout << ans;
return 0;
}
|
F
两颗树状数组，一个求比他小的个数，一个求比他小的个数之和
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| #include<bits/stdc++.h>
#define int int64_t
using namespace std;
struct Fenwick {
vector<int> tree;
int size;
explicit Fenwick(int size) : size(size) {
tree.resize(size + 1, 0);
}
void update(int pos, int value) {
while (pos <= size) {
tree[pos] += value;
pos += pos & -pos;
}
}
int query(int left, int right) {
return query(right) - query(left - 1);
}
int query(int pos) {
int res = 0;
while (pos > 0) {
res += tree[pos];
pos -= pos & -pos;
}
return res;
}
};
void solve() {
int n;
cin >> n;
vector<int> a(n);
vector<int> b(n);
map<int, int> map;
for (int i = 0; i < n; ++i)
{
cin >> a[i];
b[i] = a[i];
}
sort(b.begin(), b.end());
b.erase(unique(b.begin(), b.end()), b.end());
for (int i = 0; i < b.size(); ++i)
{
map[b[i]] = i + 1;
}
Fenwick sum(b.size());
Fenwick count(b.size());
int result = 0;
for (int i = n - 1; i >= 0; --i)
{
int hz = map[a[i]];
int sum_g = sum.query(hz + 1, b.size());
int cnt_g = count.query(hz + 1, b.size());
result += sum_g - a[i] * cnt_g;
sum.update(hz, a[i]);
count.update(hz, 1);
}
cout << result << "\n";
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
solve();
return 0;
}
|
