poj 3468(线段树)

题意：一个数字序列，Q a b查询a到b的数字和，C a b c从a到b都加c。

题解：区间修改水题。

#include 
#include 
const int N = 100005;
long long addv[N << 2], sum[N << 2], x;
int n, q, l1, r1;

void pushup(int k, int left, int right) {
	sum[k] = sum[k * 2] + sum[k * 2 + 1];
}

void pushdown(int k, int left, int right) {
	int len = right - left + 1;
	if (addv[k]) {
		addv[k * 2] += addv[k];
		addv[k * 2 + 1] += addv[k];
		sum[k * 2] += addv[k] * (len - len / 2);
		sum[k * 2 + 1] += addv[k] * (len / 2);
		addv[k] = 0;
	}
}

void build(int k, int left, int right) {
	addv[k] = 0;
	if (left == right) {
		scanf("%lld", &sum[k]);
		return;
	}
	int mid = (left + right) / 2;
	build(k * 2, left, mid);
	build(k * 2 + 1, mid + 1, right);
	pushup(k, left, right);
}

void update(int k, int left, int right) {
	if (l1 <= left && right <= r1) {
		addv[k] += x;
		sum[k] += x * (right - left + 1);
		return;
	}
	pushdown(k, left, right);
	int mid = (left + right) / 2;
	if (l1 <= mid)
		update(k * 2, left, mid);
	if (r1 > mid)
		update(k * 2 + 1, mid + 1, right);
	pushup(k, left, right);
}

long long query(int k, int left, int right) {
	if (l1 <= left && right <= r1)
		return sum[k];
	pushdown(k, left, right);
	int mid = (left + right) / 2;
	long long res = 0;
	if (l1 <= mid)
		res += query(k * 2, left, mid);
	if (r1 > mid)
		res += query(k * 2 + 1, mid + 1, right);
	return res;
}

int main() {
	while (scanf("%d%d", &n, &q) == 2) {
		build(1, 1, n);
		char str[5];
		while (q--) {
			scanf("%s%d%d", str, &l1, &r1);
			if (str[0] == 'Q')
				printf("%lld\n", query(1, 1, n));
			else if (str[0] == 'C') {
				scanf("%lld", &x);
				getchar();
				update(1, 1, n);
			}
		}
	}
	return 0;
}