| Time Limit: 330MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:
modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.
Input
The first line of input contains an integer N. The following line contains N integers, representing the sequence A1..AN.
The third line contains an integer M. The next M lines contain the operations in following form:
0 x y: modify Ax into y (|y|<=10000).
1 x y: print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.
Output
For each query, print an integer as the problem required.
Example
Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3
Output:
6
4
-3
Hint
| Added by: | Bin Jin |
| Date: | 2007-08-03 |
| Time limit: | 0.330s |
| Source limit: | 5000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: C++ 5 |
| Resource: | own problem |
单点修改，询问区间内最大连续字段和。
@TYVJ P1427 小白逛公园
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #define lc rt<<1 8 #define rc rt<<1|1 9 using namespace std; 10 const int mxn=100010; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 int n,m; 18 int data[mxn]; 19 struct node{ 20 int mx; 21 int ml,mr; 22 int smm; 23 }t[mxn<<2],tmp0; 24 void push_up(int l,int r,int rt){ 25 t[rt].smm=t[lc].smm+t[rc].smm; 26 t[rt].mx=max(t[lc].mx,t[rc].mx); 27 t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx); 28 t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml); 29 t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr); 30 return; 31 } 32 void Build(int l,int r,int rt){ 33 if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;} 34 int mid=(l+r)>>1; 35 Build(l,mid,lc); 36 Build(mid+1,r,rc); 37 push_up(l,r,rt); 38 return; 39 } 40 void change(int p,int v,int l,int r,int rt){ 41 if(l==r){ 42 if(p==l){t[rt].ml=t[rt].mr=t[rt].mx=t[rt].smm=v;} 43 return; 44 } 45 int mid=(l+r)>>1; 46 if(p<=mid)change(p,v,l,mid,lc); 47 else change(p,v,mid+1,r,rc); 48 push_up(l,r,rt); 49 return; 50 } 51 node query(int L,int R,int l,int r,int rt){ 52 // printf("%d %d %d %d %d ",L,R,l,r,rt); 53 if(L<=l && r<=R){return t[rt];} 54 int mid=(l+r)>>1; 55 node res1; 56 if(L<=mid)res1=query(L,R,l,mid,lc); 57 else res1=tmp0; 58 node res2; 59 if(R>mid)res2=query(L,R,mid+1,r,rc); 60 else res2=tmp0; 61 node res={0}; 62 res.smm=res1.smm+res2.smm; 63 res.mx=max(res1.mx,res2.mx); 64 res.mx=max(res.mx,res1.mr+res2.ml); 65 res.ml=max(res1.ml,res1.smm+res2.ml); 66 res.mr=max(res2.mr,res2.smm+res1.mr); 67 return res; 68 } 69 int main(){ 70 n=read(); 71 int i,j,x,y,k; 72 for(i=1;i<=n;i++)data[i]=read(); 73 Build(1,n,1); 74 tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0; 75 m=read(); 76 for(i=1;i<=m;i++){ 77 k=read();x=read();y=read(); 78 if(k){ 79 if(x>y)swap(x,y); 80 printf("%d ",query(x,y,1,n,1).mx); 81 } 82 else change(x,y,1,n,1); 83 } 84 return 0; 85 }