http://codevs.cn/problem/1080/
#include <vector>
#include <iostream>
#include <string.h> using namespace std; const int MAXN = 100000; struct Line {
int left, right;
int n;
}; Line tree[MAXN * 3]; void buildtr(int left, int right, int k) {
tree[k].left = left;
tree[k].right = right;
tree[k].n = 0;
if (left == right) return;
int mid = (left + right) / 2;
buildtr(left, mid, k * 2);
buildtr(mid + 1, right, k * 2 + 1);
} int query(int x, int k) {
if (tree[k].left == x && tree[k].right == x) {
return tree[k].n;
}
int mid = (tree[k].left + tree[k].right) / 2;
if (x <= mid) return query(x, k * 2);
else return query(x, k * 2 + 1);
} int query(int l, int r, int k) {
if (tree[k].left == l && tree[k].right == r) {
//cout << "k:" << k << ",l" << l << "r," << r << ",n:" << tree[k].n << endl;
return tree[k].n;
}
int mid = (tree[k].left + tree[k].right) / 2;
if (r <= mid) {
return query(l, r, k * 2);
} else if (l > mid) {
return query(l, r, k * 2 + 1);
} else {
return query(l, mid, k * 2) + query(mid + 1, r, k * 2 + 1);
}
} int update(int x, int y, int k) {
int diff = 0;
if (tree[k].left == x && tree[k].right == x) {
diff = y - tree[k].n;
tree[k].n = y;
return diff;
}
int mid = (tree[k].left + tree[k].right) / 2;
if (x <= mid) {
diff = update(x, y, k * 2);
} else {
diff = update(x, y, k * 2 + 1);
}
tree[k].n += diff;
return diff;
} int add(int l, int r, int x, int k) {
int diff = 0;
if (tree[k].left == tree[k].right) {
diff = x;
tree[k].n += x;
//cout << "diff:" << x << endl;
return x;
}
int mid = (tree[k].left + tree[k].right) / 2;
if (mid >= r) {
diff = add(l, r, x, k * 2);
} else if (mid < l) {
diff = add(l, r, x, k * 2 + 1);
} else {
diff += add(l, mid, x, k * 2);
diff += add(mid + 1, r, x, k * 2 + 1);
}
tree[k].n += diff;
return diff;
} int main() {
int n;
cin >> n;
memset(tree, sizeof(tree), 0);
buildtr(1, n, 1);
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
update(i, x, 1);
}
int m;
cin >> m;
while (m--) {
int c, x, y;
cin >> c >> x >> y;
if (c == 1) {
add(x, x, y, 1);
} else if (c == 2) {
int res = query(x, y, 1);
cout << res << endl;
}
}
}