10 16模拟赛复盘
预计分数: 100 + 40 + 0 + 20 =160
实际分数: 100 + 20 + 0 + 20 = 140
收获or反思?
1.T2与正解唯一得距离就是思考的方向,有了正解想法后,一直想用画图的方法优化,发现巨大麻烦,一直没敢写式子,要多方面尝试
2.空间运算能力太差,T4没有仔细思考
3.再认真一点!
flandre
费用提前计算? 按
a
a
a从小到大
s
o
r
t
sort
sort之后,一个数对答案的贡献为
b
i
g
g
e
r
[
a
]
∗
k
+
a
+
F
i
n
d
(
a
−
1
)
bigger[a]*k+a+Find(a-1)
bigger[a]∗k+a+Find(a−1)若贡献为正,加入
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int MAX = 1e6 + 70;
LL n, k, ans = 0, sum = 0;
LL tree[2 * MAX];
int num[MAX];
inline int read() {
int x = 0; int f = 1;
char c; c = getchar();
while(c > '9' || c < '0') { if(c == '-') f = -1; c = getchar(); }
while(c >= '0' && c <= '9') { x = x * 10 + (f * (int)(c - '0')); c = getchar(); }
return x;
}
stack<int> s;
vector<int> ANS;
inline int lowbit(int x) { return x & (-x); }
void add(int x, int val) { for(int i = x; i <= 2 * MAX; i += lowbit(i)) tree[i] += val;}
LL Find(int x) { LL res = 0; for(int i = x; i; i -= lowbit(i)) res += tree[i]; return res; }
struct made {
int val;
int id;
}a[MAX];
inline bool mycmp(made X, made Y) { return X.val < Y.val; }
int main() {
n = read(), k = read();
for(int i = 1; i <= n; i++) {
a[i].val = read();
a[i].id = i;
}
sort(a + 1, a + 1 + n, mycmp);
for(int i = 1; i <= n; i++) {
while(!s.empty() && a[i].val > a[s.top()].val) {
num[s.top()] = (n - i + 1);
s.pop();
}
s.push(i);
}
while(!s.empty()) {
num[s.top()] = 0;
s.pop();
}
for(int i = 1; i <= n; i++) {
LL Now = (1LL * a[i].val) + Find(MAX + a[i].val - 1) + (1LL * num[i] * k);
if(Now > 0) {
ans = ans + (1LL * a[i].val) + Find(MAX + a[i].val - 1);
ANS.push_back(a[i].id);
add(a[i].val + MAX, k);
}
}
printf("%lld %d\n", ans, ANS.size());
for(auto v : ANS) {
printf("%d ", v);
}
return 0;
}
meirin
简单数学题?
首先不考虑增加操作,如果单求
∑
l
=
1
n
∑
r
=
l
n
(
∑
j
=
l
r
a
i
)
(
∑
j
=
l
r
b
i
)
\sum_{l=1}^{n} \sum_{r=l}^{n}(\sum_{j=l}^{r}a_i)(\sum_{j=l}^{r}b_i)
∑l=1n∑r=ln(∑j=lrai)(∑j=lrbi)
考虑将里面的式子用前缀和表示出来即
∑ l = 1 n ∑ r = l n ( S a r − S a l − 1 ) ( S b r − S b l − 1 ) \sum_{l=1}^{n} \sum_{r=l}^{n}(Sa_r-Sa_{l-1})(Sb_r-Sb_{l-1}) ∑l=1n∑r=ln(Sar−Sal−1)(Sbr−Sbl−1)
展开
∑
l
=
1
n
∑
r
=
l
n
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S
a
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S
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S
a
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S
b
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S
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l
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)
\sum_{l=1}^{n} \sum_{r=l}^{n}(Sa_r*Sb_r+Sa_{l-1}*Sb_{l-1}-Sa_r*Sb_{l-1}-Sb{r}*Sa_{l-1})
∑l=1n∑r=ln(Sar∗Sbr+Sal−1∗Sbl−1−Sar∗Sbl−1−Sbr∗Sal−1)
令
S
a
b
i
S_{ab_i}
Sabi表示
S
a
i
∗
S
b
i
Sa_i*Sb_i
Sai∗Sbi,
S
S
a
SSa
SSa表示
S
a
Sa
Sa的前缀和,
S
S
b
SSb
SSb表示
S
b
Sb
Sb的前缀和
将式子中的
r
r
r累加起来 化简为
∑
i
=
1
n
S
a
b
i
∗
n
−
S
a
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S
S
b
n
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S
S
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)
−
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b
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∗
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S
S
a
n
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S
a
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1
)
\sum_{i=1}^{n}S_{ab_i}*n-Sa_{i-1}*(SSb_{n}-SSb_{i-1})-Sb_{i-1}*(SSa_{n}-SSa_{i-1})
∑i=1nSabi∗n−Sai−1∗(SSbn−SSbi−1)−Sbi−1∗(SSan−SSai−1)
这样就得到了一个
N
Q
NQ
NQ 时间复杂的的代码,考虑如果有修改,考虑修改的贡献
显然题目中只对
b
b
b进行操作,考虑每个
b
b
b对应的
a
a
a区间和
∑
l
=
1
i
∑
r
=
i
n
∑
j
=
l
r
a
i
\sum_{l=1}^{i}\sum_{r=i}^{n}\sum_{j=l}^{r}a_i
∑l=1i∑r=in∑j=lrai
前缀优化一维
∑
l
=
i
i
∑
r
=
i
n
S
a
r
−
S
a
l
−
1
\sum_{l=i}^{i}\sum_{r=i}^{n}S_{a_r}-S_{a_{l-1}}
∑l=ii∑r=inSar−Sal−1
将式子拆开,分别积掉一个
∑
\sum
∑后再相加
∑
i
=
1
n
i
∗
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S
S
a
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a
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−
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n
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)
(
S
S
a
i
−
1
)
\sum_{i=1}^{n}i*(SS_{a_n}-SS_{a_{i-1}})-(n-i+1)(SS_{a_{i-1}})
∑i=1ni∗(SSan−SSai−1)−(n−i+1)(SSai−1)
再对这个式子求前缀和即可
O
1
O1
O1计算贡献,总时间复杂度
O
(
n
+
q
)
O(n+q)
O(n+q)
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define lll 1LL
const int MAX = 5e5 + 70;
const int MOD = 1e9 + 7;
int n, q;
int a[MAX], b[MAX], sa[MAX], ssa[MAX];
int g[MAX], sg[MAX];
LL ans;
int main() {
scanf("%d%d", &n, &q);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
for(int i = 1; i <= n; i++) sa[i] = (1LL * sa[i - 1] + 1LL * a[i] ) %MOD;
for(int i = 1; i <= n; i++) ssa[i] = (1LL * ssa[i - 1] + 1LL * sa[i]) % MOD;
for(int i = 1; i <= n; i++) {
g[i] = ((1LL * i * ((1LL * ssa[n] - 1LL * ssa[i - 1] + MOD) % MOD) - (1LL * (n - i + 1) * (1LL * ssa[i - 1])) ) + MOD ) % MOD;
}
for(int i = 1; i <= n; i++) sg[i] = (1LL * sg[i - 1] + g[i] + MOD) % MOD;
for(int i = 1; i <= n; i++) ans = (ans + (1LL * g[i] * b[i] % MOD) ) % MOD;
for(int i = 1; i <= q; i++) {
int l, r, k; scanf("%d%d%d", &l, &r, &k);
ans = (ans + ((1LL * sg[r] - sg[l - 1] + MOD) % MOD * 1LL * k) % MOD + MOD) % MOD;
printf("%lld\n", ans);
}
return 0;
}
sakuya
最讨厌期望什么的了
首先不考虑修改
分析题意,要求期望难走程度,期望=概率*值
将值的贡献拆分成若干点对的贡献,发现每个点对在序列中相邻的概率是相同的
题意变为 ∑ l = 1 m ∑ r = 1 m d i s ( l , r ) ( l ≠ r ) ∗ P \sum_{l=1}^{m}\sum_{r=1}^{m}dis(l,r)(l \ne r)*P ∑l=1m∑r=1mdis(l,r)(l=r)∗P
首先考虑 P P P怎么计算, 发现 P P P实际等于 2 ∗ ( m − 1 ) A m − 2 m − 2 A m m \frac{2*(m-1)A_{m-2}^{m-2}}{A_{m}^{m}} Amm2∗(m−1)Am−2m−2
接下来问题转化为点对之间的贡献如何计算,发现不好处理,考虑与上面相同的处理方式,将点对贡献转化为边权*出现次数
发现出现次数可以用
f
[
v
]
∗
(
m
−
f
[
v
]
)
f[v]*(m-f[v])
f[v]∗(m−f[v])计算得出(
f
[
i
]
表示以
i
为根的子树中特殊点的个数
f[i]表示以i为根的子树中特殊点的个数
f[i]表示以i为根的子树中特殊点的个数)
重新加回修改的限制
考虑对一个点的相连边增加 k k k对答案的增量为什么?为相连边出现次数 n u m num num× k k k
上述所有操作都可以在一次树形DP中处理,时间复杂度 O ( N + Q ) O(N+Q) O(N+Q)
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define LL long long
const int MAX = 5e5 + 70;
const int MOD = 998244353;
int tot, head[MAX], n, m, q, p[MAX], fa[MAX], f[MAX]; //表示以i为子树的特殊点的个数
LL ans = 0;
int Num[MAX], sum[MAX];
vector<int> son[MAX];
vector<int> ce[MAX];
struct node { int u, v, val; }E[2 * MAX];
struct made { int l, t, id, val; }edge[MAX * 2];
void add(int u, int v, int id, int val) {
edge[++tot].l = head[u];
edge[tot].t = v;
edge[tot].val = val;
edge[tot].id = id;
head[u] = tot;
}
void dfs_pre(int x, int Fa) {
fa[x] = Fa;
f[x] = p[x];
for(int i = head[x]; i; i = edge[i].l) {
int to = edge[i].t;
if(to == Fa) continue;
ce[x].push_back(to);
son[x].push_back(to);
dfs_pre(to, x);
f[x] += f[to];
}
}
LL quick_mi(int x, int y) {
LL xx = x, res = 1;
while(y) {
if(y % 2) res = res * xx % MOD;
xx = xx * xx % MOD;
y /= 2;
}
return res % MOD;
}
signed main() {
scanf("%lld%lld", &n, &m);
for(int i = 1; i < n; i++) {
int u, v, val; scanf("%lld%lld%lld", &u, &v, &val);
E[i].u = u, E[i].v = v; E[i].val = val;
add(u, v, i, val); add(v, u, i, val);
}
for(int i = 1; i <= m; i++) {
int x; scanf("%lld", &x);
p[x] = 1;
}
LL P_up = 1, P_down = 1, P;
for(int i = 1; i <= m - 2; i++) P_up = (P_up * 1LL * i )% MOD;
for(int i = 1; i <= m; i++) P_down = (P_down * 1LL * i) % MOD;
P = 2LL * (m - 1) * P_up % MOD * quick_mi(P_down, MOD - 2) % MOD;
dfs_pre(1, 0); //儿子节点, 父亲节点,相邻的边
for(int i = 1; i < n; i++) {
if(E[i].u != fa[E[i].v]) swap(E[i].u, E[i].v);
Num[E[i].v] = (f[E[i].v] * (m - f[E[i].v])) % MOD;
}
for(int i = 1; i < n; i++) {
if(E[i].u != fa[E[i].v]) swap(E[i].u, E[i].v);
ans = (ans + (E[i].val * Num[E[i].v] % MOD) ) % MOD;
}
for(int i = 1; i <= n; i++) {
sum[i] = (sum[i] + Num[i]) % MOD;
for(int j = 0; j < ce[i].size(); j++) sum[i] = (sum[i] + Num[ce[i][j]]) % MOD;
}
scanf("%lld", &q);
for(int i = 1; i <= q; i++) {
int x, k; scanf("%lld%lld", &x, &k);
ans = (ans + (1LL * sum[x] * k)) % MOD;
printf("%lld\n", (ans * P) % MOD);
}
return 0;
}