洛谷 3373 线段树

传送门

思路：

关键在于乘与加的先后计算关系，(x + y) * k = x * k + y * k，从这里可以看出来，把加法转化为乘法计算，取消了+与*先后顺序

pushdown时，即为乘法标记 * 原有数据 + 加法标记 * 长度。

注意点：

1. 这个题数据范围取long long
2. 读入的k也是long long，传入函数时用long long
3. 个人wa点：pushdown时，暂存mul和add时，用了int

#include 
using namespace std;
#define Ln(x) (x << 1)
#define Rn(x) (x << 1) | 1
#define endl "\n"
#define ll long long
const int maxn = 1e5+5;
struct Tree{
	int l, r;
	ll data, mul, add;
	Tree(int l=0, int r=0, ll data=0):
		l(l), r(r), data(data) {mul = 1, add = 0;};
}tree[maxn*4+10];
ll a[maxn];
ll mod;

void pushup(int p){
	tree[p].data = (tree[Ln(p)].data + tree[Rn(p)].data) % mod;
}
void build(int l, int r, int p){
	tree[p] = Tree(l, r, 0);
	if(l == r){
        tree[p] = Tree(l, r, a[l]);
        return;
	}
	int mid = (l + r) >> 1;
	build(l, mid, Ln(p));
	build(mid+1, r, Rn(p));
	pushup(p);
}
void pushdown(int p){
	ll mul = tree[p].mul;
	ll add = tree[p].add;
	tree[Ln(p)].mul = (mul * tree[Ln(p)].mul) % mod;
	tree[Ln(p)].add = (tree[Ln(p)].add * mul + add) % mod;
	tree[Rn(p)].mul = (mul * tree[Rn(p)].mul) % mod;
	tree[Rn(p)].add = (tree[Rn(p)].add * mul + add) % mod;
    //乘法优先
	tree[Ln(p)].data = (tree[Ln(p)].data * mul + add * (tree[Ln(p)].r - tree[Ln(p)].l + 1)) % mod;
	tree[Rn(p)].data = (tree[Rn(p)].data * mul + add * (tree[Rn(p)].r - tree[Rn(p)].l + 1)) % mod;

	tree[p].mul = 1;
    tree[p].add = 0;
}
void update(int l, int r, int p, int op, ll d){
	int nl = tree[p].l, nr = tree[p].r;
	if(nl >= l && nr <= r){
        if(op == 1){
            tree[p].mul = tree[p].mul * d % mod;
            tree[p].data = tree[p].data * d % mod;
            tree[p].add = tree[p].add * d % mod;  //乘法优先，惩罚转化为加法
        }
        else{
            tree[p].add = (tree[p].add + d) % mod;
            tree[p].data = (d * (nr - nl + 1) + tree[p].data) % mod;
        }
		return;
	}

    if(tree[p].mul != 1 || tree[p].add) pushdown(p);

	int mid = (nl+nr) >> 1;
	if (l <= mid)	update(l, r, Ln(p), op, d);
	if (r > mid)	update(l, r, Rn(p), op, d);
	pushup(p);
}

ll query(int l, int r, int p){
	int nl = tree[p].l, nr = tree[p].r;
	if(nl >= l && nr <= r)
		return tree[p].data;
//cout << "#" << tree[1].lazy << " " << p << endl;
	if(tree[p].mul != 1 || tree[p].add) pushdown(p);
	int mid = (nl + nr) >> 1;
	ll res = 0;
	if(l <= mid)	res = (res + query(l, r, Ln(p))) % mod;
	if(r > mid) 	res = (res + query(l, r, Rn(p))) % mod;
	return res;
}

int main()
{
    int  n, m;
    cin >> n >> m >> mod;
    for (int i=1; i<=n; ++i)
        cin >> a[i];
    build(1, n, 1);
    while(m --){
        int op;
        cin >> op;
        if(op != 3){
            int l, r;
            ll d;
            cin >> l >> r >> d;
            update(l, r, 1, op, d);
        }
        else{
            int l, r;
            cin >> l >> r;
            cout << query(l, r, 1) << endl;
        }
    }
    return 0;
}