boj 451 田田的算数题 区间更新树状数组

题目链接：http://code.bupt.edu.cn/problem/p/451/

区间加上一个等差数列

这里涉及i的二次项

具体的做法见上一篇博文

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const ull B = 9973;
const int maxn = 110000;
const ll M = 23333;

ll bit0[maxn];
ll bit1[maxn];
ll bit2[maxn];

int n, q;

ll sum(ll *b, int i)
{
    ll s = 0;
    while(i > 0)
    {
        s += b[i];
        i -= i & -i;
    }
    return s;
}

void add(ll *b, int i, ll v)
{
    while(i <= n)
    {
        b[i] += v;
        i += i & -i;
    }
}

int main()
{
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
#endif

    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &q);
        memset(bit0, 0, sizeof(bit0));
        memset(bit1, 0, sizeof(bit1));
        memset(bit2, 0, sizeof(bit2));

        for(int i = 1; i <= n; i++)
        {
            int t;
            scanf("%d", &t);
            add(bit0, i, (ll)2*t);
        }

        for(int i = 0; i < q; i++)
        {
            int mode;
            scanf("%d", &mode);
            if(mode == 1)
            {
                int l, r, x, d;
                scanf("%d%d%d%d", &l, &r, &x, &d);
                add(bit0, l, (ll)(-(ll)d*(l-1)*(l-1)-((ll)(2*x+d-2*l*d)*(l-1))));
                add(bit1, l, (ll)(2*x+d-2*l*d));
                add(bit2, l, (ll)d);
                add(bit0, r+1, (ll)((ll)d*r*r+(ll)(2*x+d-2*l*d)*r));
                add(bit1, r+1, (ll)(-(2*x+d-2*l*d)));
                add(bit2, r+1, (ll)(-d));
            }
            else
            {
                int l, r;
                scanf("%d%d", &l, &r);
                ll res = 0;
                res += sum(bit0, r) + sum(bit1, r)*r + sum(bit2, r)*r*r;
                res -= sum(bit0, l-1) + sum(bit1, l-1)*(l-1) + sum(bit2, l-1)*(l-1)*(l-1);
                printf("%lld\n", res/2);
            }
        }

    }

    return 0;
}