題目鏈接：

https://nanti.jisuanke.com/t/31458

題解：

建立兩個樹狀數組，第一個是，a[1]*n+a[2]*(n-1)....+a[n]*1;第二個是正常的a[1],a[2],a[3]...a[n]

#include "bits/stdc++.h"
using namespace std;
#define ll long long
const int MAXN=1e5+10;
ll sum[MAXN],ans[MAXN];
ll num[MAXN];
ll n,q;
int lowbit(int x)
{
    return x&(-x);
}
void update(int i , ll x)
{
    ll t=x*(n-i+1);
    while(i<=n)
    {
        sum[i]+=x;
        ans[i]+=t;
        i+=lowbit(i);
    }
}
ll query1(int x)
{
    ll Sum=0;
    while(x)
    {
        Sum+=sum[x];
        x-=lowbit(x);
    }
    return Sum;
}
ll query2(int x)
{
    ll Sum=0;
    while(x)
    {
        Sum+=ans[x];
        x-=lowbit(x);
    }
    return Sum;
}
int main()
{
    while(scanf("%lld%lld",&n,&q)!=EOF)
    {

        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&num[i]);
            update(i,num[i]);
        }
        while(q--)
        {
            int a,b,c;
            scanf("%d",&a);
            if(a==1)
            {
                scanf("%d%d",&b,&c);
                ll A1=query1(c)-query1(b-1);
                ll A2=query2(c)-query2(b-1);
                printf("%lld\n",A2-A1*((n-b+1)-(c-b+1)));
            } else
            {
                scanf("%d%d",&b,&c);
                ll k=c-num[b];
                num[b]=c;
                update(b,k);
            }
        }
    }
    return 0;
}

• 262144K

Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <$x$, $y$>. If $x^i$= $x^j$ and $y^i$ = $y^j$, then <x_i

So if cat features are moving, we can think the cat is moving. If feature <$a$, $b$> is appeared in continuous frames, it will form features movement. For example, feature <$a$ , $b$ > is appeared in frame $2,3,4,7,8$, then it forms two features movement 2-3-4

Now given the features in each frames, the number of features may be different, Morgana wants to find the longest features movement.

Input

First line contains one integer $T(1\le T\le 10)$ , giving the test cases.

Then the first line of each cases contains one integer $n$ (number of frames),

In The next $n$ lines, each line contains one integer $k^i$ ( the number of features) and $2k^i$ intergers describe $k^i$features in ith frame.(The first two integers describe the first feature, the $3$rd and $4$th integer describe the second feature, and so on).

In each test case the sum number of features $N$ will satisfy $N\le 100000$ .

Output

For each cases, output one line with one integers represents the longest length of features movement.

樣例輸入復制

1
8
2 1 1 2 2
2 1 1 1 4
2 1 1 2 2
2 2 2 1 4
0
0
1 1 1
1 1 1

樣例輸出復制

3

2018徐州網絡賽H. Ryuji doesn't want to study