法一：暴力$O({n^2})$看脸过
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int a[50002],sum[50002]; 5 int main(){ 6 int n,k; 7 cin>>n>>k; 8 for(int i=1;i<=n;i++){ cin>>a[i];sum[i]=sum[i-1]+a[i];} 9 bool flag=false; 10 for(int i=0;i<n;i++){ 11 for(int j=i+1;j<=n;j++){ 12 if(sum[j]-sum[i]==k){ 13 cout<<i+1<<" "<<j<<endl; 14 flag=true; 15 break; 16 } 17 } 18 if(flag) break; 19 } 20 if(!flag) cout<<"No Solution "; 21 return 0; 22 }
法二：map优化，存储sum数组的下标，复杂度$O(nlog n)$
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 ll a[50002],s[50002]; 5 int main(){ 6 map<ll,int>m; 7 ll n,k; 8 cin>>n>>k; 9 for(ll i=1;i<=n;i++){ cin>>a[i];s[i]=s[i-1]+a[i];} 10 for(ll i=n;i>=1;i--) m[s[i]]=i; 11 bool ok=false; 12 ll l,r; 13 m[0]=0; 14 for(ll i=1;i<=n;i++){ 15 16 if(s[i]-k==0&&!ok) l=1,r=i,ok=true; 17 18 else if(m[s[i]-k]){ 19 if(m[s[i]-k]+1>i) continue; 20 if(!ok) l=m[s[i]-k]+1,r=i,ok=true; 21 else{ 22 if(m[s[i]-k]+1<l) l=m[s[i]-k]+1,r=i; 23 else if(m[s[i]-k]+1==l) r=min(r,i); 24 } 25 } 26 } 27 if(!ok) cout<<"No Solution "; 28 else cout<<l<<" "<<r<<endl; 29 30 return 0; 31 }
法三：二分复杂度$O(nlog n)$，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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 struct node{ 5 ll sum,idx; 6 }a[10002]; 7 ll b[10002]; 8 ll n,k; 9 bool cmp(node &a,node &b){ 10 return (a.sum<b.sum)||(a.sum==b.sum&&a.idx<b.idx); 11 } 12 ll erfen(int p,int id){ 13 ll l=1,r=n; 14 while(l<r){ 15 ll mid=(l+r)>>1; 16 if(a[mid].sum>=p) r=mid; 17 else l=mid+1; 18 } 19 return r; 20 } 21 int main(){ 22 cin>>n>>k; 23 for(ll i=1;i<=n;i++){ 24 cin>>a[i].sum; 25 a[i].sum+=a[i-1].sum; 26 a[i].idx=i; 27 b[i]=a[i].sum; 28 } 29 bool flag=false; 30 sort(a+1,a+n+1,cmp); 31 for(ll i=1;i<=n;i++){ 32 ll p=b[i-1]+k; 33 ll idx=erfen(p,i); 34 while(a[idx].idx<i) idx++; 35 if(a[idx].sum==p){ 36 cout<<i<<" "<<a[idx].idx<<endl; 37 flag=true; 38 break; 39 } 40 } 41 if(!flag) cout<<"No Solution "; 42 return 0; 43 44 }