hdu 4348 To the moon

题意：n个数 m次操作
操作分别为
C l r d: 把区间[l, r] 加 d
Q l r : 查询区间[l, r]的和
H l r t: 查询时间t的时候[l, r]的和
B t: 回到时间t

思路：主席树区间修改，区间求和

const int maxn = 100000 + 10;
const int maxnode = 25 * maxn;

int n, m;

struct Node
{
    int l, r;
    LL sum, add;
} tr[maxnode];
int tail;
LL a[maxn];
LL qL, qR, value;
LL sum;

#define lson tr[o].l,L,M
#define rson tr[o].r,M+1,R
#define self o,L,R
#define lc tr[o].l
#define rc tr[o].r

void build(int o, int L, int R)
{
    if (L == R) 
    {
        tr[o].sum = a[L];
        return;
    }
    tr[o].l = ++tail;
    tr[o].r = ++tail;
    int M = L + (R - L) / 2;
    build(lson);
    build(rson);
    tr[o].sum = tr[lc].sum + tr[rc].sum;
}

void maintain(int o, int L, int R)
{
    if (L < R)
    {
        tr[o].sum = tr[lc].sum + tr[rc].sum;
    }
    else tr[o].sum = a[L];
    tr[o].sum += tr[o].add * (R - L + 1);
}

void update(int& o, int L, int R)
{
    tr[++tail] = tr[o];
    o = tail;
    if (qL <= L && R <= qR)
    {
        tr[o].add += value;
    }
    else
    {
        int M = L + (R - L) / 2;
        if (qL <= M) update(lson);
        if (qR > M) update(rson);
    }
    maintain(self);
}

void query(int o, int L, int R, int add)
{
    if (qL <= L && R <= qR)
    {
        sum += tr[o].sum + add * (R - L + 1);
        return;
    }
    int M = L + (R - L) / 2;
    if (qL <= M) query(lson, add + tr[o].add);
    if (qR > M)  query(rson, add + tr[o].add);
}

int root[maxn];//表示第i个线段树的root是什么，显然root[0]=0

void init()
{
    memset(tr, 0, sizeof(tr));
    tail = 0;
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
    }
    root[0] = 0;
    build(0, 1, n);
}

void solve()
{
    char op[5];
    int t;
    int i = 0;
    while (m--) 
    {
        op[0] = 0;
        scanf("%s", op);
        if (op[0] == 'Q')
        {
            scanf("%lld%lld", &qL, &qR);
            sum = 0;
            query(root[i], 1, n, 0);
            printf("%lld
", sum);

        }
        else if (op[0] == 'C')
        {
            i++;
            root[i] = root[i-1];
            scanf("%lld%lld%lld", &qL, &qR, &value);
            update(root[i], 1, n);
        }
        else if (op[0] == 'H')
        {
            scanf("%lld%lld%d", &qL, &qR, &t);
            sum = 0;
            query(root[t], 1, n, 0);
            printf("%lld
", sum);
        }
        else 
        {
            scanf("%d", &t);
            i = t;
        }
    }
}

int main()
{
    while (scanf("%d%d", &n, &m) == 2)
    {
        init();
        solve();
    }
    return 0;
}