第1题代码丢了，论分类整理的重要性！

按回忆补一下第1题

#include <bits/stdc++.h>

using namespace std;
using ll=long long;

ll n,ans=1e9;
map<int,set<int>> mp;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1,x;i<=n;++i){
    	cin>>x;
    	mp[x].insert(i);
    }
    for(auto [_,aa]:mp){
    	if(aa.size()<2)continue;
    	ll lst=-1;
    	for(auto it:aa){
    		if(lst==-1)lst=it;
    		else{
    			ans=min(ans,it-lst);
    			lst=it;
    		}
    	}
    }
    if(ans==1e9)ans=0;
    cout<<ans;
    return 0;
}

第2题

#include <bits/stdc++.h>
using namespace std;

using ll=long long;

char s[(int)1e5+9];
ll cnt[(int)1e5+9],sz,ed[(int)1e5+9],res;
int main()
{
    scanf("%s",s+1);
    sz=strlen(s+1);
    for(int i=1;i<=sz;++i){
    	if(s[i]==s[i-1])cnt[i]=cnt[i-1]+1;
    	else cnt[i]=1;
    }
    ed[sz]=sz,ed[sz+1]=sz+1;
    for(int i=sz-1;i;--i){
    	if(s[i]==s[i+1])ed[i]=ed[i+1];
    	else ed[i]=i;
    }
    for(int i=ed[1];i<=sz;i=ed[i+1]){
    	res=res+cnt[i]*cnt[i];
    }
    ll ans=res;

    auto sol2=[](int x){
    	ll tmp=res-cnt[x]*cnt[x]-cnt[ed[x+1]]*cnt[ed[x+1]];
    	return tmp+(cnt[x]-1ll)*(cnt[x]-1ll)+(cnt[ed[x+1]]+1ll)*(cnt[ed[x+1]]+1ll);
    };
    auto sol3=[](int x){
    	ll tmp=res-cnt[x-1]*cnt[x-1]-cnt[ed[x]]*cnt[ed[x]];
    	return tmp+(cnt[x-1]+1ll)*(cnt[x-1]+1ll)+(cnt[ed[x]]-1ll)*(cnt[ed[x]]-1ll);
    };



    auto sol1=[](int x){
    	if(s[x-1]==s[x+1]){
    		ll tmp=res-cnt[x-1]*cnt[x-1]-cnt[ed[x+1]]*cnt[ed[x+1]]-1ll;
    		return tmp+(cnt[x-1]+cnt[ed[x+1]]+1ll)*(cnt[x-1]+cnt[ed[x+1]]+1ll);
    	}else{
    		ll tmp=res-cnt[x-1]*cnt[x-1]-cnt[ed[x+1]]*cnt[ed[x+1]]-1ll;
    		return tmp+max((cnt[x-1]+1)*(cnt[x-1]+1)+cnt[ed[x+1]]*cnt[ed[x+1]],cnt[x-1]*cnt[x-1]+(cnt[ed[x+1]]+1ll)*(cnt[ed[x+1]]+1ll));
    	}
    };

    for(int i=1;i<=sz;++i){
    	if(ed[i]==i){
    		ans=max(ans,sol1(i));
    	}
    	else{
    		ans=max(ans,sol2(ed[i]));
    		if(i>1)ans=max(ans,sol3(i));
    	}

    	i=ed[i];
    }
    cout<<ans<<"\n";
    return 0;
}

第3题

#include <bits/stdc++.h>
using namespace std;

using ll=long long;

const int mod=1e9+7;
std::vector<int> v;


ll n,k,ans;

map<int,int> cnt;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    for(int i=1,x;i<=n;++i){
    	cin>>x;
    	v.emplace_back(x);
    }
    sort(v.begin(),v.end());
    v.resize(unique(v.begin(),v.end())-v.begin());
    for(auto it:v){
    	ans+=upper_bound(v.begin(),v.end(),k-it)-v.begin();
    	if(2ll*it<=k)ans-=1;
    }
    ans/=2;
    ans%=mod;
    cout<<ans;
    return 0;
}

第4题

#include <bits/stdc++.h>
using namespace std;


using ll=long long;

ll n,x,a[100],k,cost;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>x;
    for(int i=1;i<=n;++i)cin>>a[i];
    cin>>k;
	for(int i=1,x;i<=k;++i){
		cin>>x;
		cost+=a[x],a[x]=0ll;
	}
	if(cost>=x){
		cout<<cost<<"\n";
	}else{
		bitset<(int)20*100000+99> vis;
		vis[0]=1;
		for(int i=1;i<=n;++i)vis|=(vis<<a[i]);
		ll cz=x-cost;
		for(int i=cz;i<=20*100000+98;++i)if(vis[i]){
			cout<<i+cost<<"\n";
			return 0;
		}
	}
    return 0;
}