POJ-3468-A Simple Problem with Integers

线段树区间更新，区间求和

#include <iostream> 
#include <cstdio> 
#define maxn 100005
#define INF 1e9
using namespace std;
typedef long long ll;

ll sum[maxn<<2], add[maxn<<2];
void Build(int n, int l, int r){

    add[n] = 0;
    if(l == r){
        scanf("%lld ", sum+n);
        return ;
    }
    int mid = (l + r) >> 1;
    Build(n<<1, l, mid);
    Build(n<<1|1, mid+1, r);
    sum[n] = sum[n<<1] + sum[n<<1|1];
} 
void Pushdown(int n, int m){

    if(add[n]){
        sum[n<<1] += (m - (m >> 1)) * add[n];
        sum[n<<1|1] += (m >> 1) * add[n];
        add[n<<1] += add[n];
        add[n<<1|1] += add[n];
        add[n] = 0;
    }
}
void Update(int n, int L, int R, int l, int r, int d){

    if(l == L && R == r){

        add[n] += d;
        sum[n] += (r - l + 1) * d;
        return ;
    }
    Pushdown(n, R - L + 1);
    int mid = (L + R) >> 1;
    if(r <= mid)
     Update(n<<1, L, mid, l, r, d);
    else if(l > mid)
     Update(n<<1|1, mid+1, R, l, r, d);
    else{
        Update(n<<1, L, mid, l, mid, d);
        Update(n<<1|1, mid+1, R, mid+1, r, d);
    } 
    sum[n] = sum[n<<1] + sum[n<<1|1];
}
void Query(int n, int L, int R, int l, int r, ll &s){

    if(L == l && R == r){
        s += sum[n];
        return ;
    }
    Pushdown(n, R - L + 1);
    int mid = (L + R) >> 1;
    if(r <= mid)
     Query(n<<1, L, mid, l, r, s);
    else if(l > mid)
          Query(n<<1|1, mid+1, R, l, r, s);
    else{
        Query(n<<1, L, mid, l, mid, s);
        Query(n<<1|1, mid+1, R, mid+1, r, s);
    }
}
int main(){

// freopen("in.txt", "r", stdin);
    int n, q;

    while(scanf("%d%d ", &n, &q) == 2){

        char ch;
        int a, b, c;
        Build(1, 1, n);
        for(int i = 0; i < q; i++){

            scanf("%c", &ch);
            if(ch == 'Q'){
                ll s = 0;
                scanf("%d%d ", &a, &b);
                Query(1, 1, n, a, b, s);
                printf("%lld\n", s);
            }
            else{
                scanf("%d%d%d ", &a, &b, &c);
                Update(1, 1, n, a, b, c);
            }
        }
    }
    return 0;
}