POJ 3468 A Simple Problem with Integers

\(POJ\) \(3468\) \(A\) \(Simple\) \(Problem\) \(with\) \(Integers\)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;

const int N = 2e5 + 10;
LL tr1[N], tr2[N];
int n, m, a[N];

#define lowbit(x) x & (-x)
void add(int x, int c) {
    for (int i = x; i < N; i += lowbit(i))
        tr1[i] += c, tr2[i] += (LL)x * c;
}

LL sum(int x) {
    LL res = 0;
    for (int i = x; i; i -= lowbit(i))
        res += (LL)(x + 1) * tr1[i] - tr2[i];
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("POJ3468.in", "r", stdin);
#endif
    scanf("%d %d", &n, &m);
    char op[110];
    int x, y, d;

    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        add(i, a[i] - a[i - 1]); // 保存基底是差分数组的树状数组
    }

    while (m--) {
        scanf("%s %d %d", op, &x, &y);
        if (op[0] == 'C') {
            scanf("%d", &d);
            add(x, d), add(y + 1, -d);
        } else
            printf("%lld\n", sum(y) - sum(x - 1));
    }
    return 0;
}