链接：https://www.nowcoder.com/acm/contest/105/H
来源：牛客网
n个桶按顺序排列，我们用1~n给桶标号。有两种操作：
1 l r c 区间[l,r]中的每个桶中都放入一个颜色为c的球 (1≤l,r ≤n,l≤r,0≤c≤60)
2 l r 查询区间[l,r]的桶中有多少种不同颜色的球 (1≤l,r ≤n,l≤r)
ac代码：
#define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<cstdio> #include<string> #include<stack> #include<ctime> #include<list> #include<set> #include<map> #include<queue> #include<vector> #include<sstream> #include<iostream> #include<functional> #include<algorithm> #include<memory.h> using namespace std; #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) #define rep(i,t,n) for(int i =(t);i<=(n);++i) #define per(i,n,t) for(int i =(n);i>=(t);--i) #define mp make_pair #define pb push_back #define mmm(a,b) memset(a,b,sizeof(a)) typedef long long ll; const int maxn = 1e5 + 5; ll tree[maxn * 4], lazy[maxn * 4]; void pushdown(ll root) { tree[root << 1] |= lazy[root]; tree[root << 1 | 1] |= lazy[root]; lazy[root << 1] |= lazy[root]; lazy[root << 1 | 1] |= lazy[root]; lazy[root] = 0; } void update(ll l, ll r, ll L, ll R, ll root, ll clr) { if (l <= L&&r >= R) { tree[root] |= clr; lazy[root] |= clr; return; } ll mid = L + R >> 1; if (lazy[root])pushdown(root); if (r <= mid)update(l, r, L, mid, root << 1, clr); else if (l > mid)update(l, r, mid+1, R, root << 1 | 1, clr); else { update(l, mid, L, mid, root << 1, clr); update(mid + 1, r, mid + 1, R, root << 1 | 1, clr); } tree[root] = tree[root << 1] | tree[root << 1 | 1]; } ll query(ll l, ll r, ll L, ll R, ll root, ll clr) { if (l <= L&&r >= R) { return tree[root]; } ll mid = L + R >> 1; if (lazy[root])pushdown(root); if (r <= mid)return query(l, r, L, mid, root << 1, clr); else if (l > mid)return query(l, r, mid+1 , R, root << 1 | 1, clr); else { return query(l, mid, L, mid, root << 1, clr)| query(mid + 1, r, mid + 1, R, root << 1 | 1, clr);; } tree[root] = tree[root << 1] | tree[root << 1 | 1]; } int main() { ios::sync_with_stdio(false); int n, m; while (cin >> n >> m) { mmm(tree, 0); mmm(lazy, 0); rep(i, 1, m) { ll x, l, r, c; cin >> x; if (x == 1) { cin >> l >> r >> c; update(l, r, 1, n, 1, 1ll << c); } if (x == 2) { cin >> l >> r; ll ans = query(l, r, 1, n,1, 1); int res = 0; while (ans) { if (ans & 1)res++; ans >>= 1; } cout << res << endl; } } } }
tle代码
#define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<cstdio> #include<string> #include<stack> #include<ctime> #include<list> #include<set> #include<map> #include<queue> #include<vector> #include<sstream> #include<iostream> #include<functional> #include<algorithm> #include<memory.h> //#define INF 0x3f3f3f3f #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) #define rep(i,t,n) for(int i =(t);i<=(n);++i) #define per(i,n,t) for(int i =(n);i>=(t);--i) #define mp make_pair #define pb push_back #define mmm(a,b) memset(a,b,sizeof(a)) //std::ios::sync_with_stdio(false); using namespace std; typedef long long ll; typedef unsigned long long ull; void smain(); #define ONLINE_JUDGE int main() { // ios::sync_with_stdio(false); #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif smain(); #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; } #define _CRT_SECURE_NO_WARNINGS const int maxn = 1e5 + 5; int a[maxn], n, q, l, r, val; typedef long long ll; struct node { int l, r; long long sum, lazy; void update(long long x) { //sum += (ll)1 * (r - l + 1)*x; //lazy += x; sum |= x; lazy |= x; } }tree[maxn * 4]; void push_up(int x) { //tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum; tree[x].sum = tree[x << 1].sum | tree[x << 1 | 1].sum; } void push_down(int x) { int lazyval = tree[x].lazy; if (lazyval) { tree[x << 1].update(lazyval); tree[x << 1 | 1].update(lazyval); tree->lazy = 0; } } void build(int x, int l, int r) { tree[x].l = l; tree[x].r = r; tree[x].sum = tree[x].lazy = 0; if (l == r) { tree[x].sum = 0; } else { int mid = l + r >> 1; build(x << 1, l, mid); build(x << 1 | 1, mid + 1, r); push_up(x); } } void update(int x, int l, int r, long long val) { int L = tree[x].l, R = tree[x].r; if (l <= L&&r >= R) { tree[x].update(val); } else { push_down(x); int mid = L + R >> 1; if (mid >= l)update(x << 1, l, r, val); if (r > mid)update(x << 1 | 1, l, r, val); push_up(x); } } long long query(int x, int l, int r) { int L = tree[x].l, R = tree[x].r; if (l <= L&&r >= R) { return tree[x].sum; } else { push_down(x); long long ans = 0; int mid = L + R >> 1; if (mid >= l)ans |= query(x << 1, l, r); if (r > mid)ans |= query(x << 1 | 1, l, r); push_up(x); return ans; } } void Run() { } void smain() { cin >> n >> q; { build(1, 1, n); rep(i, 1, q) { int x; scanf("%d", &x); //cin >> x; if (x == 1) { int l, r, c; scanf("%d%d%d", &l, &r,&c); //cin >> l >> r >> c; update(1, l, r, 1 << c); } else { int l, r; scanf("%d%d", &l, &r); //cin >> l >> r; ll res= query(1, l, r); int cnt=0; while (res) { if (res & 1)cnt++; res >>= 1; } printf("%d ", cnt); //cout << cnt << endl; } } } }