题目链接：https://cn.vjudge.net/problem/POJ-3468

题意：给你n个数，m个查询

有两种查询，一种是区间修改，给一个区间每个数都加上c

                      一种是区间查询，求这个区间的和。

需要lazy标记

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
const int maxn = 1e6 + 5;
#define ll long long

struct node
{
    ll lazy, val, cnt;
}e[maxn];
ll a[maxn];
void pushup(int cur){
    e[cur].val = e[cur<<1].val + e[cur<<1|1].val;
    e[cur].cnt = e[cur<<1].cnt + e[cur<<1|1].cnt;
}
void build(int l, int r, int cur) {
    e[cur].lazy= 0;
    if(l == r) {
        e[cur].cnt = 1;
        e[cur].val = a[l];
        return ;
    }
    int m = l + r >> 1;
    build(l, m, cur<<1);
    build(m + 1, r, cur<<1|1);
    pushup(cur);
}
void pushdown(int cur) {
    if(e[cur].lazy != 0) {
        e[cur<<1].lazy += e[cur].lazy;
        e[cur<<1|1].lazy += e[cur].lazy;
        e[cur<<1].val += e[cur].lazy * e[cur<<1].cnt;
        e[cur<<1|1].val += e[cur].lazy * e[cur<<1|1].cnt;
        e[cur].lazy = 0;
    }
}
void update(int cl, int cr, int cur, int l, int r, ll val) {
     if(l <= cl && cr <= r)
     {
         e[cur].val += val * e[cur].cnt;
         e[cur].lazy += val;
         return ;
     }
     pushdown(cur);
     int m = cl + cr >> 1;
     if(l <= m)update(cl, m, cur<<1, l, r, val);
     if(m < r)update(m + 1, cr, cur<<1|1, l, r, val);
     pushup(cur);
}
ll query(int cl, int cr, int cur, int l, int r) {
    if(l <= cl && cr <= r)
        return e[cur].val;
    pushdown(cur);
    ll ans = 0;
    int m = cl + cr >> 1;
    if(l <= m) ans += query(cl, m, cur << 1, l, r);
    if(m < r) ans += query(m + 1, cr, cur << 1 | 1, l, r);
    return ans;
}
int main()
{
    int n, l, r, m;
    while(~scanf("%d%d", &n, &m)) {
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
        }
        build(1, n, 1);
        char s[2];
        while(m--) {
            scanf("%s%d%d", s, &l, &r);
            if(s[0] == 'Q') {
                printf("%lld\n", query(1, n, 1, l, r));
            }
            else {
                ll c;
                scanf("%lld", &c);
                update(1, n, 1, l, r, c);
            }
        }


    }
    return 0;
}