第45届ICPC亚洲赛区南京站

F Fireworks

链接

https://ac.nowcoder.com/acm/contest/10272/F

题意

你想要放一个的烟花,你可以花费时间n来制作一个烟花,花费时间m点燃所有的烟花,烟花被点燃以后就释放了,但是他只有$\frac{p}{10^4}$的概率完美释放,你想完美释放至少一个烟花,那么需要的最少时间的期望是多少?

T组输入

数据范围

$T<10^4, n<10^9, m<10^9, p<10^4$

题解

假设准备了k个烟花,然后释放,这个期望值是$\frac{kn+m}{1-(1-\frac{p}{10^4})^k}$ , 这个应该只有一个极值点,也就是最小值,队友用三分过了,当然这个不好证明,其实直接使用模拟退火就可以了。下面提供了三分和模拟退火的算法。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std;
#define ll long long

const double eps = 1e-9;

double cal(double mid, double pp, ll n, ll m) {
return (n * mid + m) / (1.0 - pow(pp, mid));
}

int main() {
int t;
cin >> t;
while (t--) {
ll n, m, p;
scanf("%lld%lld%lld", &n, &m, &p);
double pp = 1.0 - 1.0 * p / 10000.0;

double t = 1e18, alpha = 0.95, x = 1.0, y = cal(x, pp, n, m), step = 0.8;
while (t > 1) {
double x_1 = x + t;
double x_2 = max(1.0, x - t);
double y_1 = cal(x_1, pp, n, m);
double y_2 = cal(x_2, pp, n, m);
double minx = x_2, miny = y_2;
if (y_1 < y_2) {
minx = x_1;
miny = y_1;
}
if (y > miny) {
x = x + step * (minx - x);
y = cal(x, pp, n, m);
}
t *= alpha;
}
double ans = 1e308;
for (ll i = max(1ll, (ll) x - 100); i <= x + 100; ++i) {
ans = min(ans, cal(i, pp, n, m));
}
printf("%.10lf\n", ans);
}
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <cmath>
#include <algorithm>

using namespace std;
#define ll long long

const double eps = 1e-9;

double cal(double mid, double pp, ll n, ll m) {
return (n * mid + m) / (1.0 - pow(pp, mid));
}

int main() {
int t; cin >> t;
while(t--) {
int n, m, p;
scanf("%d%d%d", &n, &m, &p);
double pp = 1.0 - 1.0 * p / 10000.0;
ll l = 1, r = 1e18;
while(l + 2 < r) {
ll lmid = l + (r - l) / 3;
ll rmid = l + (r - l) / 3 * 2;
if(cal(lmid, pp, n, m) >= cal(rmid, pp, n, m)) {
l = lmid;
} else if(cal(lmid, pp, n, m) < cal(rmid, pp, n, m)) {
r = rmid;
}
}
vector<double>v;
for(ll i = l; i <= r; ++i) {
v.push_back(cal(i, pp, n, m));
}
v.push_back(cal(1, pp, n, m));
sort(v.begin(), v.end());
printf("%.10lf\n", v[0]);
}
}

J Just Another Game of Stones

链接

https://ac.nowcoder.com/acm/contest/10272/J

题意

给一个长度为n的序列

操作1: 选择一段区间,把小于等于m的值变为m

操作2: 把这段区间的值取出来,放入一个列表中,向列表中加入一个值m,对列表进行nim博弈,问先手如果赢,他第一步的策略有多少种。

数据范围

$n<2e5,3S, 262MB$

题解

根据博弈论把第二个操作转化为:

问这段区间中有多少个数,他的二进制中第k位是1.

直接按位建立线段树,然后使用吉司机线段树来维护

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <cmath>
#include <algorithm>

using namespace std;

#define lson (u<<1)
#define rson (lson|1)
#define lmid ((l+r)>>1)
#define rmid (lmid+1)

int qu = 0;
int up = 0;

const int maxn = 2e5 + 100;
int a[maxn];
int mn1[maxn << 2], mn2[maxn << 2];
int lzy[maxn << 2], bit[maxn << 2][32];
int cnt[maxn << 2], xrr[maxn << 2];

void pushup(int u, int l, int r) {

map<int, int> ma;
ma[mn1[lson]] += cnt[lson];
ma[mn1[rson]] += cnt[rson];

if (lmid - l + 1 != cnt[lson]) {
ma[mn2[lson]]++;
}
if (r - rmid + 1 != cnt[rson]) {
ma[mn2[rson]]++;
}

if (ma.size() > 1) {
mn2[u] = (++ma.begin())->first;
}
cnt[u] = ma.begin()->second;

for (int i = 0; i < 31; ++i) {
bit[u][i] = bit[lson][i] + bit[rson][i];
}

mn1[u] = min(mn1[lson], mn1[rson]);
xrr[u] = xrr[lson] ^ xrr[rson];
}

void build(int u, int l, int r) {
if (l == r) {
xrr[u] = mn1[u] = mn2[u] = a[l];
lzy[u] = 0;
cnt[u] = 1;
for (int i = 0; i < 31; ++i) {
bit[u][i] = ((1 << i) & a[l]) >> i;
}
return;
}
build(lson, l, lmid);
build(rson, rmid, r);
lzy[u] = 0;
pushup(u, l, r);
}


void pushSon(int son, int l, int r, int value) {
if (mn1[son] < value) {
for (int i = 0; i < 31; ++i) {
bit[son][i] -= (((1 << i) & mn1[son]) >> i) * cnt[son];
bit[son][i] += (((1 << i) & value) >> i) * cnt[son];
}
lzy[son] = max(lzy[son], value);
if (cnt[son] & 1) {
xrr[son] ^= mn1[son];
xrr[son] ^= value;
}
mn1[son] = value;
}
}

void pushdown(int u, int l, int r) {
pushSon(lson, l, lmid, lzy[u]);
pushSon(rson, rmid, r, lzy[u]);
}

void modify(int u, int l, int r, int ql, int qr, int mx) {
up++;
if (ql <= l && r <= qr) {
if (r - l + 1 == cnt[u] || mx < mn2[u]) {
pushSon(u, l, r, mx);
return;
}
}
pushdown(u, l, r);
if (ql <= lmid) {
modify(lson, l, lmid, ql, qr, mx);
}
if (qr > lmid) {
modify(rson, rmid, r, ql, qr, mx);
}
pushup(u, l, r);
}


int query(int u, int l, int r, int ql, int qr, int bt) {
qu++;
if (ql <= l && r <= qr) {
return bit[u][bt];
}
int res = 0;
pushdown(u, l, r);
if (ql <= lmid) {
res += query(lson, l, lmid, ql, qr, bt);
}
if (qr > lmid) {
res += query(rson, rmid, r, ql, qr, bt);
}
return res;
}

int query_xr(int u, int l, int r, int ql, int qr) {
qu++;
if (ql <= l && r <= qr) {
return xrr[u];
}
int res = 0;
pushdown(u, l, r);
if (ql <= lmid) {
res ^= (query_xr(lson, l, lmid, ql, qr));
}
if (qr > lmid) {
res ^= (query_xr(rson, rmid, r, ql, qr));
}
return res;
}


int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, q;
cin >> n >> q;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
build(1, 1, n);
for (int i = 1; i <= q; ++i) {
int op, l, r, x;
cin >> op >> l >> r >> x;
if (op == 1) {
modify(1, 1, n, l, r, x);

} else {
int value = query_xr(1, 1, n, l, r) ^x;
if (value == 0) {
cout << "0\n";
continue;
}
int bt = 30;
while (1) {
if ((1 << bt) & value) {
break;
}
--bt;
}
int add = 0;
if ((value ^ x) < x) {
add = 1;
}
cout << query(1, 1, n, l, r, bt) + add << "\n";
}
}
}