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题目大意
给定一个长度为n的序列， 求m组询问C l r d 代表把区间l-r的数加上dQ l r 代表求区间l, r的和
样例
Sample Input10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4Sample Output455915
代码
#include#include #include #include using namespace std;const int N = 100010;typedef long long ll;struct Tree{ int l, r; ll sum, lazy;};Tree tree[N * 4];int a[N];int n, m;void pushdown(int node){ if(tree[node].lazy) { int l = tree[node].l; int r = tree[node].r; tree[node << 1].lazy += tree[node].lazy; tree[node << 1 | 1].lazy += tree[node].lazy; int mid = (l + r) >> 1; tree[node << 1].sum += tree[node].lazy * (mid - l + 1); tree[node << 1 | 1].sum += tree[node].lazy * (r - mid); tree[node].lazy = 0; }}void pushup(int node){ tree[node].sum = tree[node << 1].sum + tree[node << 1 | 1].sum;}void build(int node, int l, int r){ tree[node].l = l; tree[node].r = r; if(l == r) { tree[node].sum = a[l]; return; } int mid = (l + r) >> 1; build(node << 1, l, mid); build(node << 1 | 1, mid + 1, r); pushup(node);}void update(int node, int l, int r, int d){ if(tree[node].l >= l && tree[node].r <= r) { tree[node].sum += (ll)d * (tree[node].r - tree[node].l + 1); tree[node].lazy += d; return; } int x = tree[node].l; int y = tree[node].r; int mid = (x + y) >> 1; pushdown(node); if(l <= mid) update(node << 1, l, r, d); if(r > mid) update(node << 1 | 1, l, r, d); pushup(node); }ll query(int node, int l, int r){ if(tree[node].l >= l && tree[node].r <= r) return tree[node].sum; pushdown(node); int mid = (tree[node].l + tree[node].r) >> 1; ll ans = 0; if(l <= mid) ans += query(node << 1, l, r); if(r > mid) ans += query(node << 1 | 1, l, r); return ans;}int main(){ scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%d", a + i); build(1, 1, n); while(m--) { char s[2]; int l, r; scanf("%s%d%d", s, &l, &r); if(*s == 'C') { int d; scanf("%d", &d); update(1, l, r, d); } else printf("%lld\n", query(1, l, r)); } return 0;}
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