[51nod1094]和为k的连续区间

法一：暴力$O({n^2})$看脸过

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 int a[],sum[];
 int main(){
     int n,k;
     cin>>n>>k;
     for(int i=;i<=n;i++){ cin>>a[i];sum[i]=sum[i-]+a[i];}
     bool flag=false;
     for(int i=;i<n;i++){
         for(int j=i+;j<=n;j++){
             if(sum[j]-sum[i]==k){
                 cout<<i+<<" "<<j<<endl;
                 flag=true;
                 break;
             }
         }
         if(flag) break;
     }
     if(!flag) cout<<"No Solution\n";
     return ;
 }

法二：map优化，存储sum数组的下标，复杂度$O(n\log n)$

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll a[],s[];
 int main(){
     map<ll,int>m;
     ll n,k;
     cin>>n>>k;
     for(ll i=;i<=n;i++){ cin>>a[i];s[i]=s[i-]+a[i];}
     for(ll i=n;i>=;i--) m[s[i]]=i;
     bool ok=false;
     ll l,r;
     m[]=;
     for(ll i=;i<=n;i++){

         if(s[i]-k==&&!ok) l=,r=i,ok=true; 

         else if(m[s[i]-k]){
             if(m[s[i]-k]+>i) continue;
             if(!ok) l=m[s[i]-k]+,r=i,ok=true;
             else{
                 if(m[s[i]-k]+<l) l=m[s[i]-k]+,r=i;
                 else if(m[s[i]-k]+==l) r=min(r,i);
             }
         }
     }
     if(!ok) cout<<"No Solution\n";
     else cout<<l<<" "<<r<<endl;

     return ;
 }

法三：二分复杂度$O(n\log n)$，特别注意二分时的判断条件，很容易出错，idx和sum不可以一起判断，只能判断完sum，再独自判断idx，否则答案会出错

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 struct node{
     ll sum,idx;
 }a[];
 ll b[];
 ll n,k;
 bool cmp(node &a,node &b){
     return (a.sum<b.sum)||(a.sum==b.sum&&a.idx<b.idx);
 }
 ll erfen(int p,int id){
     ll l=,r=n;
     while(l<r){
         ll mid=(l+r)>>;
         if(a[mid].sum>=p) r=mid;
         else l=mid+;
     }
     return r;
 }
 int main(){
     cin>>n>>k;
     for(ll i=;i<=n;i++){
         cin>>a[i].sum;
         a[i].sum+=a[i-].sum;
         a[i].idx=i;
         b[i]=a[i].sum;
     }
     bool flag=false;
     sort(a+,a+n+,cmp);
     for(ll i=;i<=n;i++){
         ll p=b[i-]+k;
         ll idx=erfen(p,i);
         while(a[idx].idx<i) idx++;
         if(a[idx].sum==p){
             cout<<i<<" "<<a[idx].idx<<endl;
             flag=true;
             break;
         }
     }
     if(!flag) cout<<"No Solution\n";
     return ;

 }