题目链接
A. Rudolph and Cut the Rope
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
| #include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n;
int t;
int a[N],b[N];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
cin >> n;
for(int i = 1;i <= n;++i)
cin >> a[i] >> b[i];
int cnt = 0;
for(int i = 1;i <= n;++i)
{
if(b[i] < a[i])
cnt++;
}
cout << cnt << endl;
}
}
|
B. Rudolph and Tic-Tac-Toe
纯码力题
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
| #include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n;
int t;
char a[5][5];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
for(int i = 1;i <= 3;++i)
{
for(int j = 1;j <= 3;++j)
{
cin >> a[i][j];
}
}
bool success = 0;
char ans;
for(int i = 1;i <= 3;++i)
{
if(success)
break;
for(int j = 1;j <= 3;++j)
{
if(j == 3)
{
ans = a[i][j];
if(ans != '.')
success = 1;
}
if(a[i][j + 1] != a[i][j])
break;
}
}
for(int i = 1;i <= 3;++i)
{
if(success)
break;
for(int j = 1;j <= 3;++j)
{
if(j == 3)
{
ans = a[j][i];
if(ans != '.')
success = 1;
}
if(a[j + 1][i] != a[j][i])
break;
}
}
for(int i = 1;i <= 3;++i)
{
if(success)
break;
if(i == 3)
{
ans = a[i][i];
if(ans != '.')
success = 1;
}
if(a[i][i] != a[i + 1][i + 1])
{
break;
}
}
for(int i = 1;i <= 3;++i)
{
if(success)
break;
if(i == 3)
{
ans = a[i][4 - i];
if(ans != '.')
success = 1;
}
if(a[i][4 - i] != a[i + 1][3 - i])
{
break;
}
}
if(success)
{
if(ans == '.')
cout << "DRAW" << endl;
else
cout << ans << endl;
}
else
cout << "DRAW" << endl;
}
}
|
C. Rudolf and the Another Competition
简单贪心
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
| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
int h,n,m,t;
struct node
{
ll point;
ll penalty;
bool s;
}a[N];
bool cmp(node x,node y)
{
if(x.point != y.point)
{
return x.point > y.point;
}else if(x.penalty != y.penalty)
{
return x.penalty < y.penalty;
}else
{
return x.s > y.s;
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
cin >> n >> m >> h;
memset(a,0,sizeof(a));
vector<vector<int> >po;
po.resize(n + 1);
for(int i = 1;i <= n;++i)
{
po[i].resize(m);
for(int j = 0;j < m;++j)
{
cin >> po[i][j];
}
}
for(int i = 1;i <= n;++i)
{
sort(po[i].begin(),po[i].end());
for(int j = 1;j < m;++j)
po[i][j] += po[i][j - 1];
int pos = 0;
while(po[i][pos] <= h && pos < m)
{
a[i].point++;
a[i].penalty += po[i][pos];
pos++;
}
if(i == 1)
a[i].s = 1;
}
sort(a + 1,a + 1 + n,cmp);
for(int i = 1;i <= n;++i)
{
if(a[i].s)
{
int pos = i;
while(a[pos].point == a[i].point && a[i].penalty == a[pos].penalty&&pos>0)
pos--;
cout << pos+1 << endl;
break;
}
}
}
}
|
D. Rudolph and Christmas Tree
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
| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
int t,n;
double d,h;
ll y[N];
int main()
{
cin >> t;
while(t--)
{
cin >> n >> d >> h;
for(int i = 1;i <= n;++i)
cin >> y[i];
double ans = 0;
for(int i = 1;i <= n;++i)
{
if(y[i + 1] - y[i] >= h || i == n)
{
ans += d * h / 2;
}else
{
double temp_h = y[i + 1] - y[i];
double upper = (1.0 - temp_h / h) * d;
ans += (upper + d) / 2 * temp_h;
}
}
printf("%lf\n",ans);
}
}
|
E. Rudolf and Snowflakes
easy version可以map暴力，hard version枚举次数后二分底数
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
| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MX = 1e18;
int t;
ll n;
void check(ll x)
{
for(int i = 2;i <= 62;++i)
{
ll l = 2,r = 1e9;
while(l < r)
{
ll mid = (l + r) >> 1;
ll sum = 0,m = 1;
bool over = 0;
for(int j = 0;j <= i;++j)
{
if(MX - sum < m)
{
over = 1;
break;
}
sum += m;
if(MX / m < mid && j < i)
{
over = 1;
break;
}
m *= mid;
}
if(sum == x)
{
cout << "YES" << endl;
return;
}else if(sum > x || over)
{
r = mid;
}else if(sum < x)
{
l = mid + 1;
}
}
}
cout << "NO" << endl;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
cin >> n;
check(n);
}
}
|
G. Rudolf and CodeVid-23
可以看作最短路模型，每种药可看作两种状态之间费用为1的通路，建完图后Dijkstra即可
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
| #include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> PII;
const int N = 1 << 10;
int dis[N];
int t,n,m;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
cin >> n >> m;
memset(dis,0x3f,sizeof(dis));
bitset<10> tmp;
cin >> tmp;
int st = (int)tmp.to_ulong();
vector<pair<PII,int> > edge;
edge.resize(m + 1);
for(int i = 1;i <= m;++i)
{
cin >> edge[i].second;
cin >> tmp;
edge[i].first.first = ((1 << n) - 1) ^ (int)(tmp.to_ulong());
cin >> tmp;
edge[i].first.second = (int)(tmp.to_ulong());
}
dis[st] = 0;
set<PII> q ;
q.insert({0,st});
while(q.size())
{
auto temp = *q.begin();
q.erase(temp);
for(int i = 1;i <= m;++i)
{
int ne = temp.second & edge[i].first.first;
ne |= edge[i].first.second;
if(dis[ne] > temp.first + edge[i].second)
{
q.erase({dis[ne],ne});
dis[ne] = temp.first + edge[i].second;
q.insert({dis[ne],ne});
}
}
}
if(dis[0] > 1e8)
cout << "-1" << '\n';
else
cout << dis[0] << '\n';
}
}
|