ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study（树状数组）

 

 

 

 

Output

For each question, output one line with one integer represent the answer.

样例输入

5 3
1 2 3 4 5
1 1 3
2 5 0
1 4 5

样例输出

10
8
两个树状数组一个维护a[i]前缀合,一个维护(n-i+1)*a[i]前缀和。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define esp 1e-9
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
typedef unsigned long long ll;
ll sum[100006],n,m;
ll pos[100005];
ll a[100005];
void add(int x,ll val){
    for(int i=x;i;i-=lowbit(i))
        sum[i]+=val;
}
ll query(int x){
    ll ans=0;
    for(int i=x;i<=100004;i+=lowbit(i))
        ans+=sum[i];
    return ans;
}
void add_one(int x,ll val){
    for(int i=x;i;i-=lowbit(i))
        pos[i]+=val;
}
ll query_one(int x){
    ll ans=0;
    for(int i=x;i<=100004;i+=lowbit(i))
        ans+=pos[i];
    return ans;
}
int main()
{
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        memset(sum,0,sizeof(sum));
        memset(pos,0,sizeof(pos));
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            add(i,1ll*a[i]*(n-i+1));
            add_one(i,a[i]);
        }
        while(m--){
            ll op,x,y;
            scanf("%lld%lld%lld",&op,&x,&y);
            if(op==1)
            {
                ll l=query(x);
                ll r=query(y+1);
                ll u=query_one(x);
                ll d=query_one(y+1);
                //printf("%lld\n",u-d);
                printf("%lld\n",l-r-(1ll*(n-y)*(u-d)));
            }
            else{
                add(x,1ll*(-1)*a[x]*(n-x+1));
                add(x,1ll*(n-x+1)*y);
                add_one(x,1ll*(-1)*a[x]);
                a[x]=y;
                add_one(x,1ll*a[x]);
            }
        }
    }
    return 0;
}