题目很模板
10127. 「一本通 4.3 练习 1」最大数
#include
using namespace std;
const int N = 1000005;
int val[N], a[N], m, p, n;
void pushup(int p) { val[p] = max(val[p * 2], val[p * 2 + 1]); }
void modify(int p, int l, int r, int pos, int v) {
if (l == r) {
val[p] = v;
} else {
int mid = (l + r) >> 1;
if (pos <= mid)
modify(p * 2, l, mid, pos, v);
else
modify(p * 2 + 1, mid + 1, r, pos, v);
pushup(p);
}
}
int query(int p, int l, int r, int ql, int qr) {
if (l > qr || r < ql)
return 0;
if (l >= ql && r <= qr)
return val[p];
int mid = (l + r) / 2;
return max(query(p * 2, l, mid, ql, qr), query(p * 2 + 1, mid + 1, r, ql, qr));
}
int main() {
cin >> m >> p;
n = m;
int len = 0;
int lastans = 0;
for (int i = 1; i <= m; i++) {
char op;
int x;
cin >> op >> x;
if (op == 'A') {
x += lastans;
x %= p;
modify(1, 1, n, ++len, x);
} else {
cout << (lastans = query(1, 1, n, len - x + 1, len)) << endl;
}
}
} 10128. 「一本通 4.3 练习 2」花神游历各国
维护区间最大值及位置，修改的时候如果当前区间最大值大于1就继续，否则剪枝。这样等价于做了不超过O(nlogn)次单点修改，实现上和普通线段树略有出入。
#include
using namespace std;
#define ll long long
const ll N = 400005;
ll n, m, val[N], mv[N], mx[N], a[N];
void pushup(ll p) {
mv[p] = max(mv[p * 2], mv[p * 2 + 1]);
if (mx[p] == mx[p * 2])
mx[p] = mx[p * 2];
else
mx[p] = mx[p * 2 + 1];
val[p] = val[p * 2] + val[p * 2 + 1];
}
void build(ll p, ll l, ll r) {
if (l == r) {
val[p] = mv[p] = a[l];
mx[p] = l;
} else {
build(p * 2, l, (l + r) / 2);
build(p * 2 + 1, (l + r) / 2 + 1, r);
pushup(p);
}
}
void modify(ll p, ll l, ll r, ll ql, ll qr) {
if (l > qr || r < ql || mv[p] <= 1)
return;
if (l == r) {
val[p] = mv[p] = sqrt(val[p]);
return;
}
modify(p * 2, l, (l + r) / 2, ql, qr);
modify(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr);
pushup(p);
}
ll query(ll p, ll l, ll r, ll ql, ll qr) {
if (l > qr || r < ql)
return 0;
if (l >= ql && r <= qr)
return val[p];
return query(p * 2, l, (l + r) / 2, ql, qr) + query(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr);
}
int main() {
cin >> n;
for (ll i = 1; i <= n; i++) cin >> a[i];
build(1, 1, n);
cin >> m;
for (ll i = 1; i <= m; i++) {
ll t1, t2, t3;
cin >> t1 >> t2 >> t3;
if (t1 == 1) {
cout << query(1, 1, n, t2, t3) << endl;
} else {
modify(1, 1, n, t2, t3);
}
}
} 10129. 「一本通 4.3 练习 3」维护序列
#include
using namespace std;
typedef long long ll;
const ll N = 1000005;
ll val[N], ta[N], tb[N], n, P, m, a[N], t1, t2, t3, t4;
void pushup(ll p) { val[p] = (val[p * 2] + val[p * 2 + 1]) % P; }
void pushdown(ll p, ll l, ll r) {
if (tb[p] != 1) {
tb[p * 2] *= tb[p];
tb[p * 2 + 1] *= tb[p];
ta[p * 2] *= tb[p];
ta[p * 2 + 1] *= tb[p];
val[p * 2] *= tb[p];
val[p * 2 + 1] *= tb[p];
tb[p] = 1;
}
if (ta[p]) {
ta[p * 2] += ta[p];
ta[p * 2 + 1] += ta[p];
val[p * 2] += ta[p] * ((l + r) / 2 - l + 1);
val[p * 2 + 1] += ta[p] * (r - (l + r) / 2);
ta[p] = 0;
}
val[p * 2] %= P;
val[p * 2 + 1] %= P;
ta[p * 2] %= P;
ta[p * 2 + 1] %= P;
tb[p * 2] %= P;
tb[p * 2 + 1] %= P;
}
void build(ll p, ll l, ll r) {
ta[p] = 0;
tb[p] = 1;
if (l == r) {
val[p] = a[l];
return;
}
build(p * 2, l, (l + r) / 2);
build(p * 2 + 1, (l + r) / 2 + 1, r);
pushup(p);
}
void modify(ll p, ll l, ll r, ll ql, ll qr, ll add, ll mul) {
if (l > qr || r < ql)
return;
if (l >= ql && r <= qr) {
ta[p] += add;
tb[p] *= mul;
ta[p] *= mul;
val[p] *= mul;
val[p] += add * (r - l + 1);
ta[p] %= P;
tb[p] %= P;
val[p] %= P;
} else {
pushdown(p, l, r);
modify(p * 2, l, (l + r) / 2, ql, qr, add, mul);
modify(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr, add, mul);
pushup(p);
}
}
ll query(ll p, ll l, ll r, ll ql, ll qr) {
if (l > qr || r < ql)
return 0;
if (l >= ql && r <= qr)
return val[p];
pushdown(p, l, r);
return (query(p * 2, l, (l + r) / 2, ql, qr) + query(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr)) % P;
}
int main() {
ios::sync_with_stdio(false);
cin >> n >> P;
for (ll i = 1; i <= n; i++) cin >> a[i];
build(1, 1, n);
cin >> m;
for (ll i = 1; i <= m; i++) {
cin >> t1 >> t2 >> t3;
if (t1 <= 2)
cin >> t4;
if (t1 == 1) {
modify(1, 1, n, t2, t3, 0, t4);
}
if (t1 == 2) {
modify(1, 1, n, t2, t3, t4, 1);
}
if (t1 == 3) {
cout << query(1, 1, n, t2, t3) << endl;
}
}
}