AtCoder Beginner Contest 378
A - Pairing
思路
根据题意分类模拟判断即可
AC代码
#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 2e5+10, MAX = 1e9, INF = -1e9;
int e;
map<int,int> mp;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
for(int i=1;i<=4;i++)
{
cin>>e;
mp[e]++;
}
for(auto i:mp){
if(i.second==4){
cout<<2<<endl;
return 0;
}
if(i.second==3){
cout<<1<<endl;
return 0;
}
if(i.second==2&&mp.size()==2){
cout<<2<<endl;
return 0;
}
if(i.second==2&&mp.size()==3){
cout<<1<<endl;
return 0;
}
if(mp.size()==4){
cout<<0<<endl;
return 0;
}
}
return 0;
}
B - Garbage Collection
思路
对于每一个查询，找到第一个大于等于目标天数的值，暴力找会TLE，二分即可；
AC代码
#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 105, MAX = 1e9, INF = -1e9;
int n;
int a[N];
int d[N];
int q;
int t,s;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
cin>>d[i];
}
cin>>q;
while(q--)
{
cin>>t>>s;
int l=0;int r=MAX;
while(l<r)
{
int mid=(l+r)>>1;
if(mid*a[t]+d[t]>=s)r=mid;
else l=mid+1;
}
cout<<l*a[t]+d[t]<<endl;
}
return 0;
}
C - Repeating
思路
对于每个字符，建立索引数组，对于每次构造，二分查找目标，时间复杂度\(O(nlogn)\);
AC代码
#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 2e5+10, MAX = 1e9, INF = -1e9;
int n;
int a[N];
int b[N];
map<int,vector<int>> mp;
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
mp[a[i]].pb(i);
}
for(int i=1;i<=n;i++){
int l=0;int r=mp[a[i]].size()-1;
while(l<r)
{
int mid=(l+r+1)>>1;
if(mp[a[i]][mid]<i)l=mid;
else r=mid-1;
}
if(mp[a[i]][l]!=i)cout<<mp[a[i]][l]<<" ";
else cout<<"-1"<<" ";
}
return 0;
}
D - Count Simple Paths
思路
第一次看错了，以为是大于 \(K\) 的路径都要，结果只是找等于的，数据非常小，直接DFS；
AC代码
#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 12, MAX = 1e9, INF = -1e9;
int h,w,k;
char g[N][N];
bool st[N][N];
int ans=0;
int dx[5]={0,1,-1,0,0};
int dy[5]={0,0,0,1,-1};
void dfs(int x,int y,int l)
{
if(l>=k+1){
ans++;
return ;
}
st[x][y]=true;
for(int i=1;i<=4;i++)
{
int a=x+dx[i];
int b=y+dy[i];
if(a>=1&&a<=h&&b>=1&&b<=w&&g[a][b]!='#'&&!st[a][b]){
st[a][b]=true;
dfs(a,b,l+1);
st[a][b]=false;
}
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>h>>w>>k;
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
{
cin>>g[i][j];
}
}
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
{
if(g[i][j]!='#'){
memset(st,false,sizeof st);
dfs(i,j,1);
}
}
}
cout<<ans<<endl;
return 0;
}
E - Mod Sigma Problem
思路
所求式子中需要求和，想到前缀和预处理，但是无法处理 \(mod\) 运算，不妨将前缀和直接定义为模运算之后的和，即 \(S [i]=(A [1]+A [2]+\dots+A [i]) \mathbin{\mathrm{mod}} M\), 这样定义后，原式将会退化：
\[\sum_{1\leq l\leq r\leq N}\left(\left(\sum_{l\leq i\leq r} A_i\right)\mathbin{\mathbin{mod}}M \right)=
\sum_{1\leq l\leq r\leq N}(S_r-S_{l-1})\mathbin{\mathbin{mod}}M
\]
此时由于 \(S_i<M\),
\[(S_r-S_{l-1})\mathbin{\mathrm{mod}} M=S_r-S_{1-1}+ \begin{cases}0 & (S_{l-1} \leq S_r) \\M & (S_{l-1}>S_r),\end{cases}
\]
记\(X_r\)为小于时的数量，则：
\[\sum_{r=1}^N \sum_{l=1}^r(S_r-S_{l-1}) \mathbin{\mathrm{mod}} M= \sum_{r=1}^N \left(S_r\times r-
\sum_{l=1}^rS_{l-1}+ M \times X_r\right).
\]
最后用树状数组维护 \(X_r\) ,时间复杂度为\(O(nlogm)\);
AC代码
#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 2e5+10, MAX = 1e9, INF = -1e9;
int n,m,a[N],s[N],g[N],ans,tr[N];
int lowbit(int x)
{
return x&(-x);
}
void add(int x)
{
if(x==0)
{
return;
}
while(x<=m)
{
tr[x]++;
x+=lowbit(x);
}
}
int sum(int x)
{
int sum=0;
while(x>0)
{
sum+=tr[x];
x-=lowbit(x);
}
return sum;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>a[i];
a[i]%=m;
s[i]=(s[i-1]+a[i])%m;
g[i]=g[i-1]+s[i];
}
for(int i=1;i<=n;i++)
{
ans=ans+i*s[i]-g[i-1]+m*(sum(m)-sum(s[i]));
add(s[i]);
}
cout<<ans<<endl;
return 0;
}
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