Master of Sequence（二分+树状数组）

题目描述
There are two sequences a1,a2,…,an , b1,b2,…,bn . Let . There are m operations within three kinds as following:
• 1 x y: change value ax to y.
• 2 x y: change value bx to y.
• 3 k: ask min{t|k≤S(t)}

输入
The first line contains a integer T (1≤T≤5) representing the number of test cases.
For each test case, the first line contains two integers n (1≤n≤100000), m (1≤m≤10000).
The following line contains n integers representing the sequence a1,a2,…,an .
The following line contains n integers representing the sequence b1,b2,…,bn .
The following m lines, each line contains two or three integers representing an operation mentioned above.
It is guaranteed that, at any time, we have 1≤ai≤1000, 1≤bi,k≤109 . And the number of queries (type 3 operation) in each test case will not exceed 1000.

输出
For each query operation (type 3 operation), print the answer in one line.

样例输入
2
4 6
2 4 6 8
1 3 5 7
1 2 3
2 3 3
3 15
1 3 8
3 90
3 66
8 5
2 4 8 3 1 3 6 24
2 2 39 28 85 25 98 35
3 67
3 28
3 73
3 724
3 7775

样例输出
17
87
65
72
58
74
310
2875

思路
面对需要更改值，并需要求区间和的序列，可以用树状数组来维护求值，这题要求最小值，可以用二分法寻找答案
具体推理可见：
HDU 6274 Master of sequence（二分答案 + 树状数组）

代码实现

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <cstdlib>
#include <cstring>
#include <map>
#include <set>
#include <stack>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int N=1005;
const int M=100005;
int T,n,m;
int a[M],b[M],t[M];
int lowbit(int x)
{
    return x&(-x);
}

struct BinaryIndexTree
{
    int c[N];

    inline void update(int x,int y)
    {
        x++;
        for(int i=x;i<N;i+=lowbit(i)) c[i]+=y;
    }

    inline int getsum(int x)
    {
        int res=0;
        x++;
        for(int i=x;i;i-=lowbit(i)) res+=c[i];
        return res;
    }
}BIT[N];

bool check(ll x,ll y)
{
    ll s=0;
    for(int i=1;i<=1000;i++)
    {
        s+=(x/i)*t[i]-(t[i]-BIT[i].getsum(x%i));
        if(s>=y+1000-i) return true;
    }
    return s>=y;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        ll s=0;
        memset(t,0,sizeof(t));
        memset(BIT,0,sizeof(BIT));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            t[a[i]]++;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            s+=b[i]/a[i];
            BIT[a[i]].update(b[i]%a[i],1);
        }
        while(m--)
        {
            ll ans;
            int op,x,y;
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%d",&x,&y);
                s=s-b[x]/a[x]+b[x]/y;
                t[a[x]]--;
                BIT[a[x]].update(b[x]%a[x],-1);
                BIT[y].update(b[x]%y,1);
                a[x]=y;
                t[y]++;
            }
            if(op==2)
            {
                scanf("%d%d",&x,&y);
                s=s-b[x]/a[x]+y/a[x];
                BIT[a[x]].update(b[x]%a[x],-1);
                BIT[a[x]].update(y%a[x],1);
                b[x]=y;
            }
            if(op==3)
            {
                scanf("%d",&x);
                ll xx=x+s;
                ll l=0,r=2e12,mid;
                while(l<=r)
                {
                    mid=(l+r)>>1;
                    if(check(mid,xx))
                    {
                        ans=mid;
                        r=mid-1;
                    }
                    else l=mid+1;
                }
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}