[2068] Luo Valley and statistics

Title Description

Given a length of n (n <= 100000), the initial values ​​are 0 as a sequence of numbers on the modified x (x <= 10000) certain locations times, each time adding a number, and submit y (y <= 10000) issue, and seek each segment interval. One second time limit.

Input and output formats

Input formats:

 

The first line number 1, n denotes the length of the sequence

Second line number 1, w denotes the number of operations

W is sequentially behind the line, respectively and added interrogate operation

Thereto, denoted by x, y represented by the interrogation

X is the format "xa b" represents a plus b at a position of the sequence

y is in the format of "ya b" indicates a query and adding the interval b

 

Output formats:

 

One per line number, which are the result of each inquiry

 

Sample input and output

Input Sample # 1:  Copy

5 
4 
x 8 March 
y 1 3 
x April 9 
and April 3

Output Sample # 1:  Copy

8 
17 

segments Number practices:

// luogu-judger-enable-o2
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define maxn 100007 
using namespace std;
int n,m,x,y,z;
char jjj[4];
long long Sum[maxn<<2],Add[maxn<<2];
int A[maxn];

void PushUp(int rt){Sum[rt]=Sum[rt*2]+Sum[rt*2+1];}

void Build(int l,int r,int rt){
    if(l==r) {
        Sum[rt]=A[l];
        return;
    }
    int m=(l+r)>>1;
    Build(l,m,rt<<1);
    Build(m+1,r,rt<<1|1);
    PushUp(rt);
}
void PushDown(int rt,int ln,int rn){
    if(Add[rt]){
        Add[rt<<1]+=Add[rt];
        Add[rt<<1|1]+=Add[rt];
        Sum[rt<<1]+=Add[rt]*ln;
        Sum[rt<<1|1]+=Add[rt]*rn;
        Add[rt]=0;
    }
}
long long Query(int L,int R,int l,int r,int rt){
    if(L <= l && r <= R)
        return Sum[rt];
    int m=(l+r)>>1;
    PushDown(rt,m-l+1,r-m); 
    long long ANS=0;
    if(L <= m) ANS+=Query(L,R,l,m,rt<<1);
    if(R >  m) ANS+=Query(L,R,m+1,r,rt<<1|1);
    return ANS;
} 

void Update(int L,int R,int C,int l,int r,int rt){
    if(L <= l && r <= R){
        Sum[rt]+=C*(r-l+1);
        Add[rt]+=C;
        return ; 
    }
    int m=(l+r)>>1;
    PushDown(rt,m-l+1,r-m);
    if(L <= m) Update(L,R,C,l,m,rt<<1);
    if(R >  m) Update(L,R,C,m+1,r,rt<<1|1); 
    PushUp(rt);
} 

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++)
            A[i]=0;
        Build(1,n,1);
        while(m--)
        {
            scanf("%s",jjj);
            if(jjj[0]=='x') {
                scanf("%d%d",&x,&z);
                Update(x,x,z,1,n,1);
            }
            else {
                scanf("%d%d",&x,&y);
                printf("%lld\n",Query(x,y,1,n,1));
            }
        }
    }
    return 0;
}

Fenwick tree solution as follows:

’

#include<iostream>
using namespace std;
const int MAXN=100000+2;
int a[MAXN];
int n,w;
int lowbit(int x){
    return x&(-x);
}
void add(int p,int x){
    while(p<=n){
        a[p]+=x;
        p+=lowbit(p);
    }
}
int ask(int p){
    int ans=0;
    while(p>0){
        ans+=a[p];
        p-=lowbit(p);
    }
    return ans;
}
int main(){
    cin>>n>>w;
    for(int i=1;i<=w;i++){
        char q;
        int y,z;
        cin>>q>>y>>z;
        if(q=='x') add(y,z);
        else cout<<ask(z)-ask(y-1)<<endl;
    }
    return 0;
}