A题 楼兰图腾
链接：https://ac.nowcoder.com/acm/contest/1032/A
树状数组 + 逆序对
#include
using namespace std;
#define lowbit(x) (x & -x)
typedef long long ll;
const int maxn = 2e5 + 10;
int a[maxn], c[maxn], l[maxn], r[maxn], n;
void update(int p, int val) {
while (p <= n) {
c[p] += val;
p += lowbit(p);
}
}
int ask(int x) {
int res = 0;
while (x > 0) {
res += c[x];
x -= lowbit(x);
}
return res;
}
int main() {
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false), cin.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
//求正序，前面有几个比a[i]大
for (int i = 1; i <= n; i++) {
update(a[i], 1);
l[i] = ask(a[i] - 1);
}
memset(c, 0, sizeof c);
//树状数组清零求逆序，后面有几个比a[i]大
for (int i = n; i >= 1; i--) {
update(a[i], 1);
r[i] = ask(a[i] - 1);
}
ll ans = 0;
//依次枚举每个点作为中间点，以该点位中心的 ‘v’ 个数显然是 left[i] * right[i]
for (int i = 2; i <= n - 1; i++) {
ans += 1ll * (i - l[i] - 1) * (n - i - r[i]);
}
cout << ans << " ";
ans = 0;
//同理枚举 '^'的个数
for (int i = 2; i <= n - 1; i++) {
ans += 1ll * l[i] * r[i];
}
cout << ans << endl;
return 0;
}
B题 A Tiny Problem with intergers
链接：https://ac.nowcoder.com/acm/contest/1032/B
树状数组 区间修改 + 单点查询
#include
using namespace std;
#define lowbit(x) (x & -x)
typedef long long ll;
const int maxn = 1e5 + 10;
int tr[maxn], n, q, a, pre;
void add(int i, int v) {
while (i <= n) {
tr[i] += v;
i += lowbit(i);
}
}
int getsum(int x) {
int res = 0;
while (x) {
res += tr[x];
x -= lowbit(x);
}
return res;
}
int main() {
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false), cin.tie(0);
cin >> n >> q;
for (int i = 1; i <= n; ++i)cin >> a, add(i, a - pre), pre = a;
char c; int u, v, a;
while (q--) {
cin >> c >> u;
if (c == 'C') {
cin >> v >> a; add(u, a), add(v + 1, - a);
}
else
cout << getsum(u) << endl;
}
}
C题 A Simple Problem with Integers
链接：https://ac.nowcoder.com/acm/contest/1032/C
树状数组 区间修改 + 区间查询 / （线段树 or 分块）
#include
using namespace std;
#define lowbit(x) (x & -x)
typedef long long ll;
const int maxn = 1e5 + 10;
ll n, q, tr1[maxn], tr2[maxn], a, pre;
void add(int x, int v) {
for (int i = x; i <= n; i += lowbit(i)) tr1[i] += v, tr2[i] += 1ll * v * (x - 1);
}
ll getsum(int x) {
long long sum = 0;
for (int i = x; i; i -= lowbit(i)) sum += tr1[i] * x - tr2[i];
return sum;
}
int main() {
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false), cin.tie(0);
cin >> n >> q;
for (int i = 1; i <= n; i++) cin >> a, add(i, a - pre), pre = a;
char c; int u, v, a;
while (q--) {
cin >> c >> u >> v;
if (c == 'C') {
cin >> a; add(u, a), add(v + 1, - a);
}
else
cout << getsum(v) - getsum(u - 1) << endl;
}
}
D题 Lost Cows
链接：https://ac.nowcoder.com/acm/contest/1032/D
树状数组 + 二分（or倍增）
#include
using namespace std;
const int maxn = 2e5 + 10;
long long tr[maxn];
void add(int x, int v){
if (x == 0) return;
while (x < maxn) {
tr[x] += v;
x += x & -x;
}
}
long long query(int x){
long long rec = 0;
while (x) {
rec += tr[x];
x -= x & -x;
}
return rec;
}
int a[maxn], ans[maxn];
int main(){
freopen("in.txt", "r", stdin);
int n; scanf("%d", &n);
for (int i = 2; i <= n; i++)
scanf("%d", &a[i]);
//二分，查询前mid个数有多少1比较，更新区间
for (int i = n; i; i--) {
int l = 1, r = n + 1, t;
while (l <= r) {
int mid = l + r >> 1;
if (query(mid) + a[i] < mid) t = mid, r = mid - 1;
else l = mid + 1;
}
ans[i] = t;
add(t, 1);
}
for (int i = 1; i <= n; i++)
printf("%d\n", ans[i]);
}