请注意：题目背景与题目可能没有关系
第一题，性质题，找到序列的最大值与最小值,我们发现如果只有正数的话和只有负数的话都很好处理，正数正序处理类似前缀加，负数后缀加,那如果正负都有，该怎么办呢?其实我们可以吧序列全变为正的或负的吧，但是需要比较一下最大值最小值，如果都变成正的话，对被卡掉，例如1 1 1 1 1 1 1 1 -1e9就死了
点击查看代码
#include <bits/stdc++.h>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ll long long
#define pb push_back
#define ull unsigned long long
#define pii pair<int,int>
#define lid (rt<<1)
#define rid (rt<<1|1)
#define pii pair<int,int>
using namespace std;
const int N = 1e5+5;
ll n,a[N],b[N];int cnt;
vector <pii> put;
void zh(int l,int r)
{for(int i=l;i<r;i++){if(a[i]<=a[i+1])continue;a[i+1]+=a[i];put.pb({i,i+1});cnt++;}
}
void fu(int l,int r)
{for(int i=r;i>l;i--){if(a[i]>=a[i-1])continue;a[i-1]+=a[i];put.pb({i,i-1});cnt++; }
}
int main()
{speed();// freopen("in.in","r",stdin);// freopen("out.out","w",stdout);cin>>n;for(int i=1;i<=n;i++){cin>>a[i];b[i]=a[i];}sort(b+1,b+1+n);if(b[1]>=0){zh(1,n);cout<<cnt<<endl;for(auto [l,r]:put){cout<<l<<" "<<r<<endl;}}else{if(b[n]<=0){fu(1,n);cout<<cnt<<endl;for(auto [l,r]:put){cout<<l<<" "<<r<<endl;}}else{ll mx=b[n],pos=max_element(a+1,a+1+n)-a;ll mi=b[1],pp=min_element(a+1,a+1+n)-a;if(mx>-mi){for(int i=n;i;i--){int st=i;while(a[st]<=0&&st>0){a[st]+=mx;put.pb({pos,st});if(mx<a[st]){mx=a[st];pos=st;}cnt++;st--;}}// for(int i=1;i<=n;i++)cout<<a[i]<<" ";zh(1,n);// for(int i=1;i<=n;i++)cout<<a[i]<<" "cout<<cnt<<endl;for(auto [l,r]:put){cout<<l<<" "<<r<<endl;} }else{for(int i=n;i;i--){int st=i;while(a[st]>=0&&st>0){a[st]+=mi;put.pb({pp,st});if(mi>a[st]){mi=a[st];pp=st;}cnt++;st--;}}// for(int i=1;i<=n;i++)cout<<a[i]<<" ";fu(1,n);// for(int i=1;i<=n;i++)cout<<a[i]<<" "cout<<cnt<<endl;for(auto [l,r]:put){cout<<l<<" "<<r<<endl;} }}}// for(int i=1;i<n;i++)// {// if(a[i]>a[i+1])// {// cout<<i<<" "<<a[i]<<" "<<a[i+1]<<endl;// cout<<"Wa"<<endl;// return 0;// }// }return 0;
}
这题，如果直接用暴力矩阵乘法的话是\(O(n^3)\),显然过不了，但我们可以发现一个性质，就是我们可以随机一个\(n\times 1\)矩阵，值得注意的是矩阵不满足交换律，但是\(randP \times A\times B \equiv randP\times C\)矩阵由\(n^2\)转换为\(n\)
这样我们就转换为\(O(n^2)\)的复杂度了
点击查看代码
#include <bits/stdc++.h>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ll long long
#define pb push_back
#define ull unsigned long long
#define pii pair<int,int>
#define lid (rt<<1)
#define rid (rt<<1|1)
using namespace std;
const ll mod=998244353;
ll n;
mt19937 seed(random_device{}());
int rand(int l,int r)
{uniform_int_distribution<int>range(l,r);return range(seed);
}
struct Matrix
{vector <vector<ll>> a;int n,m;Matrix(int _n,int _m):n(_n),m(_m){a.resize(n+1,vector<ll>(m+1));}void resize(int n,int m){a.resize(n+1,vector<ll>(m+1));}Matrix operator * (const Matrix& A)const{Matrix ans(n,1);for(int i=1;i<=n;i++)for(int j=1;j<=A.m;j++)for(int k=1;k<=m;k++)ans.a[i][j]=(ll)(ans.a[i][j]+(ll)a[i][k]*A.a[k][j]*1ll%mod)%mod;return ans;}void T(){cout<<"******"<<endl;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)cout<<a[i][j]<<" ";cout<<endl;}}
};
int main()
{speed();// freopen("sequence2.in","r",stdin);freopen("in.in","r",stdin);freopen("out.out","w",stdout);int T;cin>>T;while(T--){cin>>n;Matrix A(n,n),B(n,n),C(n,n),b(n,1);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>A.a[i][j];for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>B.a[i][j];for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>C.a[i][j];for(int i=1;i<=n;i++)b.a[i][1]=rand(1,16);// b.T();C=C*b;// C.T();b=B*b;//这里注意啊，为什么这么写A*(B*b)=C*b// b.T();b=A*b;// b.T();bool f=1;for(int i=1;i<=n;i++){if(!f)break;for(int j=1;j<=1;j++){if(b.a[i][j]!=C.a[i][j]){f=0;break;}}}if(f)cout<<"Yes"<<endl;else cout<<"No"<<endl;}return 0;
}
这题暴力分就是二维背包,考场上这都能写挂，挂了35分
点击查看代码
#include <bits/stdc++.h>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ll long long
#define pb push_back
#define ull unsigned long long
#define pii pair<int,int>
#define lid (rt<<1)
#define rid (rt<<1|1)
using namespace std;
const int N = 1e4+5;
short n,A,B;short dp[N][N];
struct ac
{short a,b;
}q[N];
int ans;
bool vis[N];
void dfs(int a,int b,int cnt)
{if(a>A||b>B){ans=max(ans,cnt);return;}for(int i=1;i<=n;i++){if(vis[i])continue;vis[i]=1;dfs(a+q[i].a,b+q[i].b,cnt+1);vis[i]=0;}
}
int main()
{speed();// freopen("construct2.in","r",stdin);// freopen("in.in","r",stdin);// freopen("out.out","w",stdout);cin>>n>>A>>B;short mx=0;for(int i=1;i<=n;i++){cin>>q[i].a>>q[i].b;mx=max(q[i].a,mx);mx=max(q[i].b,mx);}mx=max(mx,A);mx=max(mx,B);// dfs(0,0,0);int ans=0;for(int i=1;i<=n;i++){for(int j=mx;j>=q[i].a;j--){for(int k=mx;k>=q[i].b;k--){// if(j>=A&&j-q[i].a>=A)continue;// if(k>=B&&k-q[i].b>=B)continue;dp[j][k]=max<short>(dp[j][k],dp[j-q[i].a][k-q[i].b]+1);// if(j>=A&&j-q[i].a>=A)continue;// if(k>=B&&k-q[i].b>=B)continue;ans=max<short>(ans,dp[j][k]);// cout<<j<<" "<<k<<" "<<dp[j][k]<<endl;}}}// for(int i=0;i<=mx;i++)ans=max(ans,dp[A][i]);// for(int i=0;i<=mx;i++)ans=max(ans,dp[i][B]);cout<<dp[A][B]+(dp[A][B]!=n)<<endl;return 0;
}
正解
考虑到\(O(nV^2)\)中\(V\)太大了，我们可以换一种枚举方式，换成\(f_{i,j,k}\)枚举前\(i\)人中选了\(j\)个人，\(A\)值为多少，复杂度为\(O(n^2V)\)这样就可以过了
点击查看代码
#include <bits/stdc++.h>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ll long long
#define pb push_back
#define ull unsigned long long
#define pii pair<int,int>
#define lid (rt<<1)
#define rid (rt<<1|1)
using namespace std;
const int N = 1e4+5;
int n,A,B,dp[2][85][N];
struct ac
{int a,b;
}q[N];
bool cmp(ac a,ac b)
{return a.b<b.b;
}
int ans;
bool vis[N];int main()
{speed();// freopen("construct2.in","r",stdin);// freopen("in.in","r",stdin);// freopen("out.out","w",stdout);cin>>n>>A>>B;int mx=0;for(int i=1;i<=n;i++){cin>>q[i].a>>q[i].b;mx=max(q[i].a,mx);mx=max(q[i].b,mx);}mx=max(mx,A);mx=max(mx,B);memset(dp,0x3f,sizeof dp);dp[0][0][0]=0;;for(int i=1;i<=n;i++){for(int j=0;j<=i;j++){for(int v=A;v>=0;v--){dp[i&1][j][v]=dp[(i-1)&1][j][v];if(j>=1&&v>=q[i].a)dp[i&1][j][v]=min(dp[i&1][j][v],dp[(i-1)&1][j-1][v-q[i].a]+q[i].b);// cout<<dp[i][j][v]<<endl;}}}for(int j=1;j<=n;j++)for(int v=A;v>=0;v--)if(dp[n&1][j][v]<=B)ans=max(ans,j);cout<<min(ans+1,n)<<endl;return 0;
}
