【洛谷】P4588 [TJOI2018] 数学计算 的题解 + 线段树板子代码
题解
论我的快读快写的神奇故事
划水一个线段树 + + + 离散化。
一如既往的打了一个暴力，神奇的一分不得。果然线段树的威力在这里qaq
总体思路就是将数据按时间排序，建线段树，维护区间乘。这样的话根节点就是到现在为止的所有数的乘积。
代码
#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {
inline int read() {
register int x = 0, f = 1;
register char c = getchar();
while (c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
inline void write(int x) {
if(x < 0) putchar('-'), x = -x;
if(x > 9) write(x / 10);
putchar(x % 10 + '0');
return;
}
}
using namespace fastIO;
int T, q, op, m;
ll mod, sum[1000005];
void update(int now) {
sum[now] = (sum[now << 1] * sum[now << 1|1]) % mod;
}
void build(int now, int l, int r) {
if(l == r) {
sum[now] = 1;
return;
}
int mid = (l + r) >> 1;
build(now << 1, l, mid);
build(now << 1|1, mid + 1, r);
update(now);
}
void change(int c, int l, int r, int L, int R, int p) {
if(l >= L && r <= R) {
sum[c] = p;
return;
}
int mid = (l + r) >> 1;
if(L <= mid)
change(c << 1, l, mid, L, R, p);
if(R > mid)
change(c << 1 | 1, mid + 1, r, L, R, p);
update(c);
}
int main() {
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> T;
for(int i = 1; i <= T; i ++) {
cin >> q >> mod;
build(1, 1, q);
for(int i = 1; i <= q; i ++) {
cin >> op >> m;
if(op == 1) {
change(1, 1, q, i, i, m);
sum[1] %= mod;
}
else {
change(1, 1, q, m, m, 1);
}
cout << sum[1] % mod << endl;
}
}
return 0;
}
附加
其实这就是一个有一点点像线段树板子，总体来说没有什么难度，线段树的话，其实只要写熟练了以后，基本就很简单了。
很多大佬在 NOIP 的比赛前都会打一遍线段树，可见这个的重要性了吧。
这里也给出两个线段树的板子题的代码。
【洛谷】P3372 【模板】线段树 1
代码
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
ll n, a[100005], d[270005], b[270005];
void build(ll l, ll r, ll p) {
if (l == r) {
d[p] = a[l];
return;
}
ll m = l + ((r - l) >> 1);
build(l, m, p << 1), build(m + 1, r, (p << 1) | 1);
d[p] = d[p << 1] + d[(p << 1) | 1];
}
void update(ll l, ll r, ll c, ll s, ll t, ll p) {
if (l <= s && t <= r) {
d[p] += (t - s + 1) * c, b[p] += c;
return;
}
ll m = s + ((t - s) >> 1);
if (b[p])
d[p << 1] += b[p] * (m - s + 1), d[(p << 1) | 1] += b[p] * (t - m), b[p << 1] += b[p], b[(p << 1) | 1] += b[p];
b[p] = 0;
if (l <= m)
update(l, r, c, s, m, p << 1);
if (r > m) update(l, r, c, m + 1, t, (p << 1) | 1);
d[p] = d[p << 1] + d[(p << 1) | 1];
}
ll getsum(ll l, ll r, ll s, ll t, ll p) {
if (l <= s && t <= r) return d[p];
ll m = s + ((t - s) >> 1);
if (b[p])
d[p << 1] += b[p] * (m - s + 1), d[(p << 1) | 1] += b[p] * (t - m), b[p << 1] += b[p], b[(p << 1) | 1] += b[p];
b[p] = 0;
ll sum = 0;
if (l <= m)
sum =getsum(l, r, s, m, p << 1);
if (r > m) sum += getsum(l, r, m + 1, t, (p << 1) | 1);
return sum;
}
int main() {
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
ll q, i1, i2, i3, i4;
cin >> n >> q;
for (ll i = 1; i <= n; i++) cin >> a[i];
build(1, n, 1);
while (q--) {
cin >> i1 >> i2 >> i3;
if (i1 == 2)
cout << getsum(i2, i3, 1, n, 1) << endl;
else
cin >> i4, update(i2, i3, i4, 1, n, 1);
}
return 0;
}
【洛谷】P3373 【模板】线段树 2
代码
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
inline ll read() {
register ll x = 0, f = 1;
register char c = getchar();
while (c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int n, m;
ll mod;
ll a[100005], sum[400005], mul[400005], laz[400005];
void up(int i) { sum[i] = (sum[(i << 1)] + sum[(i << 1) | 1]) % mod; }
void pd(int i, int s, int t) {
int l = (i << 1), r = (i << 1) | 1, mid = (s + t) >> 1;
if (mul[i] != 1) {
mul[l] *= mul[i];
mul[l] %= mod;
mul[r] *= mul[i];
mul[r] %= mod;
laz[l] *= mul[i];
laz[l] %= mod;
laz[r] *= mul[i];
laz[r] %= mod;
sum[l] *= mul[i];
sum[l] %= mod;
sum[r] *= mul[i];
sum[r] %= mod;
mul[i] = 1;
}
if (laz[i]) {
sum[l] += laz[i] * (mid - s + 1);
sum[l] %= mod;
sum[r] += laz[i] * (t - mid);
sum[r] %= mod;
laz[l] += laz[i];
laz[l] %= mod;
laz[r] += laz[i];
laz[r] %= mod;
laz[i] = 0;
}
return;
}
void build(int s, int t, int i) {
mul[i] = 1;
if (s == t) {
sum[i] = a[s];
return;
}
int mid = s + ((t - s) >> 1);
build(s, mid, i << 1);
build(mid + 1, t, (i << 1) | 1);
up(i);
}
void chen(int l, int r, int s, int t, int i, ll z) {
int mid = s + ((t - s) >> 1);
if (l <= s && t <= r) {
mul[i] *= z;
mul[i] %= mod;
laz[i] *= z;
laz[i] %= mod;
sum[i] *= z;
sum[i] %= mod;
return;
}
pd(i, s, t);
if (mid >= l) chen(l, r, s, mid, (i << 1), z);
if (mid + 1 <= r) chen(l, r, mid + 1, t, (i << 1) | 1, z);
up(i);
}
void add(int l, int r, int s, int t, int i, ll z) {
int mid = s + ((t - s) >> 1);
if (l <= s && t <= r) {
sum[i] += z * (t - s + 1);
sum[i] %= mod;
laz[i] += z;
laz[i] %= mod;
return;
}
pd(i, s, t);
if (mid >= l) add(l, r, s, mid, (i << 1), z);
if (mid + 1 <= r) add(l, r, mid + 1, t, (i << 1) | 1, z);
up(i);
}
ll getans(int l, int r, int s, int t,int i) {
int mid = s + ((t - s) >> 1);
ll tot = 0;
if (l <= s && t <= r) return sum[i];
pd(i, s, t);
if (mid >= l) tot += getans(l, r, s, mid, (i << 1));
tot %= mod;
if (mid + 1 <= r) tot += getans(l, r, mid + 1, t, (i << 1) | 1);
return tot % mod;
}
int main() {
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cout.tie(0);
int x, y, bh;
ll z;
n = read(), m = read(), mod = read();
for(int i = 1; i <= n; i ++) a[i] = read();
build(1, n, 1);
for(int i = 1; i <= m; i ++) {
bh = read();
if(bh == 1) {
x = read(), y = read(), z = read();
chen(x, y, 1, n, 1, z);
}
else if(bh == 2) {
x = read(), y = read(), z = read();
add(x, y, 1, n, 1, z);
}
else if(bh == 3) {
x = read(), y = read();
cout << getans(x, y, 1, n, 1) << endl;
}
}
return 0;
}
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