题目链接：uva 12436 - Rip Van Winkle‘s Code

题目大意：四种操作，操作见题目。

解题思路：即用线段树维护一个等差数列，因为一个等差加上一个等差还是一个等差数列，所以对于每个节点记录区

间左端的值，也就是首项，以及公差即可。因为还有一个S操作，所以要开一个标记记录区间值是否相同。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
const int maxn = 250100;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2];
ll nd[maxn << 2], ad[maxn << 2], s[maxn << 2];

void pushup(int u);
void pushdown (int u);

inline int length(int u) {
    return rc[u] - lc[u] + 1;
}

inline void change (int u, ll a) {
    v[u] = 1;
    ad[u] = 0;
    nd[u] = a;
    s[u] = a * length(u);
}

inline void maintain (int u, ll a, ll d) {
    if (v[u] && lc[u] != rc[u]) {
        pushdown(u);
        pushup(u);
    }

    v[u] = 0;
    nd[u] += a;
    ad[u] += d;
    ll n = length(u);
    s[u] += a * n + (((n-1) * n) / 2) * d;
}

inline void pushup (int u) {
    s[u] = s[lson(u)] + s[rson(u)];
}

inline void pushdown (int u) {
    if (v[u]) {
        change(lson(u), nd[u]);
        change(rson(u), nd[u]);
        v[u] = nd[u] = 0;
    } else if (nd[u] || ad[u]) {
        maintain(lson(u), nd[u], ad[u]);
        maintain(rson(u), nd[u] + length(lson(u)) * ad[u], ad[u]);
        nd[u] = ad[u] = 0;
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    nd[u] = ad[u] = s[u] = 0;

    if (l == r)
        return;
    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify(int u, int l, int r, ll a, ll d) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, a + d * (lc[u] - l), d);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, a, d);
    if (r > mid)
        modify(rson(u), l, r, a, d);
    pushup(u);
}

void set(int u, int l, int r, ll a) {
    if (l <= lc[u] && rc[u] <= r) {
        change(u, a);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        set(lson(u), l, r, a);
    if (r > mid)
        set(rson(u), l, r, a);
    pushup(u);
}

ll query (int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    pushdown(u);
    ll ret = 0;
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        ret += query(lson(u), l, r);
    if (r > mid) 
        ret += query(rson(u), l, r);
    pushdown(u);
    return ret;
}

int N;

int main () {
    while (~scanf("%d", &N)) {
        char op[5];
        int l, r, x;
        build(1, 1, 250000);
        while (N--) {
            scanf("%s%d%d", op, &l, &r);
            if (op[0] == ‘A‘)
                modify(1, l, r, 1, 1);
            else if (op[0] == ‘B‘)
                modify(1, l, r, r - l + 1, -1);
            else if (op[0] == ‘C‘) {
                scanf("%d", &x);
                set(1, l, r, x);
            } else
                printf("%lld\n", query(1, l, r));
        }
    }
    return 0;
}

uva 12436 - Rip Van Winkle's Code(线段树)