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;
}