A:Exam
Solved.
温暖的签。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 1e3 + 10;
6
7 int k;
8 char str1[maxn], str2[maxn];
9
10 int main()
11 {
12 while(~scanf("%d",&k))
13 {
14 scanf("%s", str1 + 1);
15 scanf("%s", str2 + 1);
16 int cnt1 = 0, cnt2 = 0;
17 int len = strlen(str1 + 1);
18 for(int i = 1; i <= len; ++i)
19 {
20 if(str1[i] == str2[i]) cnt1++;
21 else cnt2++;
22 }
23 int ans = 0;
24 if(cnt1 >= k)
25 {
26 ans = k + cnt2;
27 }
28 else
29 {
30 ans = len - (k - cnt1);
31 }
32 printf("%d\n", ans);
33 }
34 return 0;
35 }
B:Coprime Integers
Solved.
枚举gcd反演
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long ll;
6
7 const int maxn = 1e7 + 10;
8
9 bool check[maxn];
10 int prime[maxn];
11 ll mu[maxn];
12
13 void Moblus()
14 {
15 memset(check, false, sizeof check);
16 mu[1] = 1;
17 int tot = 0;
18 for(int i = 2; i < maxn; ++i)
19 {
20 if(!check[i])
21 {
22 prime[tot++] = i;
23 mu[i] = -1;
24 }
25 for(int j = 0; j < tot; ++j)
26 {
27 if(i * prime[j] > maxn) break;
28 check[i * prime[j]] = true;
29 if(i % prime[j] == 0)
30 {
31 mu[i * prime[j]] = 0;
32 break;
33 }
34 else
35 {
36 mu[i * prime[j]] = -mu[i];
37 }
38 }
39 }
40 }
41
42 ll sum[maxn];
43
44 ll calc(int n, int m)
45 {
46 ll ans = 0;
47 if(n > m) swap(n, m);
48 for(int i = 1, la = 0; i <= n; i = la + 1)
49 {
50 la = min(n / (n / i), m / (m / i));
51 ans += (ll)(sum[la] - sum[i - 1]) * (n / i) * (m / i);
52 }
53 return ans;
54 }
55
56 ll a, b, c, d;
57
58 int main()
59 {
60 Moblus();
61 for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + mu[i];
62 while(~scanf("%lld %lld %lld %lld", &a, &b, &c, &d))
63 {
64 ll ans = calc(b, d) - calc(a - 1, d) - calc(b, c - 1) + calc(a - 1, c - 1);
65 printf("%lld\n", ans);
66 }
67 return 0;
68 }
C:Contest Setting
Solved.
题意：
有n个题目，每个题目的难度值不同，要选出k个组成一套$Contest$
要求每个题目难度不同，求方案数
思路：
$把难度值相同的题目放在一起作为一种 dp[i][j] 表示选到第i种题目，已经选了k种的方案数$
$然后做01背包即可，注意加上的值为cnt[i] 表示该种题目有多少个$
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long ll;
6
7 const ll MOD = 998244353;
8 const int maxn = 1e3 + 10;
9
10 int n, k, pos;
11 ll cnt[maxn];
12 ll dp[maxn];
13 map<int, int>mp;
14
15 int main()
16 {
17 while(~scanf("%d %d", &n, &k))
18 {
19 mp.clear();
20 pos = 0;
21 memset(cnt, 0, sizeof cnt);
22 for(int i = 1; i <= n; ++i)
23 {
24 int num;
25 scanf("%d", &num);
26 if(mp[num] == 0) mp[num] = ++pos;
27 int id = mp[num];
28 cnt[id]++;
29 }
30 memset(dp, 0, sizeof dp);
31 dp[0] = 1;
32 for(int i = 1; i <= pos; ++i)
33 {
34 for(int j = k; j >= 1; --j)
35 {
36 dp[j] = (dp[j] + dp[j - 1] * cnt[i] % MOD) % MOD;
37 }
38 }
39 printf("%lld\n", dp[k]);
40 }
41 return 0;
42 }
D:Count The Bits
Solved.
题意：
在$[0, 2^b - 1] 中所有k的倍数以二进制形式表示有多少个1$
思路：
$dp[i][j] 表示枚举二进制位数，j 模k的余数， 表示的是前i位中模k的余数为j的数中1的个数$
$cnt[i][j] 表示当前二进制位为第i位， 模k的余数为j的数的个数$
直接转移即可。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long ll;
6
7 const ll MOD = (ll)1e9 + 9;
8 const int maxn = 1e3 + 10;
9
10 int k, b;
11 ll cnt[maxn][maxn];
12 ll dp[maxn][maxn];
13
14 int main()
15 {
16 while(~scanf("%d %d", &k, &b))
17 {
18 memset(cnt, 0, sizeof cnt);
19 memset(dp, 0, sizeof dp);
20 cnt[1][0]++;
21 cnt[1][1 % k]++;
22 dp[1][1 % k]++;
23 ll tmp = 1;
24 for(int i = 1; i <= b; ++i)
25 {
26 tmp = (tmp << 1) % k;
27 for(int j = 0; j < k; ++j)
28 {
29 //0
30 cnt[i + 1][j] = (cnt[i + 1][j] + cnt[i][j]) % MOD;
31 dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
32 //1
33 ll tmp2 = (tmp + j) % k;
34 cnt[i + 1][tmp2] = (cnt[i + 1][tmp2] + cnt[i][j]) % MOD;
35 dp[i + 1][tmp2] = ((dp[i + 1][tmp2] + cnt[i][j]) % MOD + dp[i][j]) % MOD;
36 }
37 }
38 printf("%lld\n", dp[b][0]);
39 }
40 return 0;
41 }
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1010
5 #define ll long long
6 const ll MOD = (ll)1e9 + 9;
7 ll dp[N][N], cnt[N][N];
8 int n, k;
9
10 int main()
11 {
12 while (scanf("%d%d", &k, &n) != EOF)
13 {
14 memset(dp, 0, sizeof dp);
15 memset(cnt, 0, sizeof cnt);
16 ++dp[1][1 % k];
17 ++cnt[1][1 % k];
18 ++cnt[1][0];
19 ll tmp = 1 % k;
20 for (int i = 2; i <= n; ++i)
21 {
22 tmp = tmp * 2 % k;
23 for (int j = 0; j < k; ++j)
24 {
25 dp[i][j] = dp[i - 1][j];
26 cnt[i][j] = cnt[i - 1][j];
27 if (j - tmp >= 0)
28 {
29 dp[i][j] = (dp[i][j] + dp[i - 1][j - tmp] + cnt[i - 1][j - tmp]) % MOD;
30 cnt[i][j] = (cnt[i][j] + cnt[i - 1][j - tmp]) % MOD;
31 }
32 else
33 {
34 dp[i][j] = (dp[i][j] + dp[i - 1][k + j - tmp] + cnt[i - 1][k + j - tmp]) % MOD;
35 cnt[i][j] = (cnt[i][j] + cnt[i - 1][k + j - tmp]) % MOD;
36 }
37 }
38 }
39 printf("%lld\n", dp[n][0]);
40 }
41 return 0;
42 }
E:Cops And Robbers
Solved.
题意：
在一个$n * m 的矩阵中，有些地方可以建墙，但是不同的墙开销不同，求最小开销把劫匪围住$
思路：
拆点最小割，但是是点权，拆点即可。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 4e3 + 10;
6 const int INF = 0x3f3f3f3f;
7
8 struct Edge{
9 int to, flow, nxt;
10 Edge(){}
11 Edge(int to, int nxt, int flow):to(to), nxt(nxt), flow(flow){}
12 }edge[maxn << 2];
13
14 int head[maxn], dep[maxn];
15 int S,T;
16 int N, n, m, tot;
17
18 void Init(int n)
19 {
20 N = n;
21 //memset(head, -1, sizeof head);
22 for(int i = 0; i < N; ++i) head[i] = -1;
23 tot = 0;
24 }
25
26 void addedge(int u, int v, int w, int rw = 0)
27 {
28 edge[tot] = Edge(v, head[u], w); head[u] = tot++;
29 edge[tot] = Edge(u, head[v], rw); head[v] = tot++;
30 }
31
32 bool BFS()
33 {
34 //memset(dep, -1, sizeof dep);
35 for(int i = 0; i < N; ++i) dep[i] = -1;
36 queue<int>q;
37 q.push(S);
38 dep[S] = 1;
39 while(!q.empty())
40 {
41 int u = q.front();
42 q.pop();
43 for(int i = head[u]; ~i; i = edge[i].nxt)
44 {
45 if(edge[i].flow && dep[edge[i].to] == -1)
46 {
47 dep[edge[i].to] = dep[u] + 1;
48 q.push(edge[i].to);
49 }
50 }
51 }
52 return dep[T] < 0 ? 0 : 1;
53 }
54
55 int DFS(int u, int f)
56 {
57 if(u == T || f == 0) return f;
58 int w, used = 0;
59 for(int i = head[u]; ~i; i = edge[i].nxt)
60 {
61 if(edge[i].flow && dep[edge[i].to] == dep[u] + 1)
62 {
63 w = DFS(edge[i].to, min(f - used, edge[i].flow));
64 edge[i].flow -= w;
65 edge[i ^ 1].flow += w;
66 used += w;
67 if(used == f) return f;
68 }
69 }
70 if(!used) dep[u] = -1;
71 return used;
72 }
73
74 int Dicnic()
75 {
76 int ans = 0;
77 while(BFS())
78 {
79 int tmp = DFS(S, INF);
80 if(tmp == INF) return -1;
81 ans += tmp;
82 }
83 return ans;
84 }
85
86 int c;
87 int C[maxn];
88 char mp[100][100];
89
90 int calc(int i, int j)
91 {
92 return i * 31 + j;
93 }
94
95 int main()
96 {
97 while(~scanf("%d %d %d", &m, &n, &c))
98 {
99 int base = n * 31 + m + 31;
100 int x, y;
101 Init(4000);
102 for(int i = 1; i <= n; ++i)
103 {
104 for(int j = 1; j <= m; ++j)
105 {
106 scanf(" %c", &mp[i][j]);
107 if(mp[i][j] == ‘B‘) { x = i, y = j; }
108 }
109 }
110 for(int i = 1; i <= c; ++i) scanf("%d", C + i);
111 S = 0, T = calc(x, y);
112 for(int i = 1; i <= n; ++i)
113 {
114 addedge(S, calc(i, 1), INF);
115 addedge(S, calc(i, m), INF);
116 }
117 for(int i = 1; i <= m; ++i)
118 {
119 addedge(S, calc(1, i), INF);
120 addedge(S, calc(n, i), INF);
121 }
122 for(int i = 1; i <= n; ++i)
123 {
124 for(int j = 1; j <= m; ++j)
125 {
126 if(i != n)
127 {
128 addedge(calc(i, j) + base, calc(i + 1, j), INF);
129 addedge(calc(i + 1, j) + base, calc(i, j), INF);
130 }
131 if(j != m)
132 {
133 addedge(calc(i, j)+ base, calc(i, j + 1), INF);
134 addedge(calc(i, j + 1) + base, calc(i, j), INF);
135 }
136 }
137 }
138 for(int i = 1; i <= n; ++i)
139 {
140 for(int j = 1; j <= m; ++j)
141 {
142 if(mp[i][j] >= ‘a‘ && mp[i][j] <= ‘z‘)
143 {
144 addedge(calc(i, j), calc(i, j) + base, C[mp[i][j] - ‘a‘ + 1]);
145 }
146 else
147 {
148 addedge(calc(i, j), calc(i, j) + base, INF);
149 }
150 }
151 }
152 int ans = Dicnic();
153 printf("%d\n", ans);
154 }
155 return 0;
156 }
F:Rectangles
Solved.
题意：
二维平面上有一些平行坐标轴的矩形，求有多少区间是被奇数个矩形覆盖。
思路：
扫描线，只是把区间加减换成区间01状态翻转，现场学扫描线可还行..
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define ll long long
5 #define N 200010
6 #define pii pair <int, int>
7 int n, mx, my, bx[N], by[N];
8 ll res;
9 struct Rec
10 {
11 int x[2], y[2];
12 void scan()
13 {
14 for (int i = 0; i < 2; ++i) scanf("%d%d", x + i, y + i);
15 if (x[0] > x[1]) swap(x[0], x[1]);
16 if (y[0] > y[1]) swap(y[0], y[1]);
17 bx[++mx] = x[0];
18 bx[++mx] = x[1];
19 by[++my] = y[0];
20 by[++my] = y[1];
21 }
22 }rec[N];
23 vector <pii> v[N];
24
25 void Hash()
26 {
27 sort(bx + 1, bx + 1 + mx);
28 sort(by + 1, by + 1 + my);
29 mx = unique(bx + 1, bx + 1 + mx) - bx - 1;
30 my = unique(by + 1, by + 1 + my) - by - 1;
31 for (int i = 1; i <= n; ++i)
32 {
33 for (int j = 0; j < 2; ++j)
34 {
35 rec[i].x[j] = lower_bound(bx + 1, bx + 1 + mx, rec[i].x[j]) - bx;
36 rec[i].y[j] = lower_bound(by + 1, by + 1 + my, rec[i].y[j]) - by;
37 }
38 }
39 }
40
41 namespace SEG
42 {
43 struct node
44 {
45 ll sum, val;
46 int lazy;
47 node () {}
48 node (ll sum, ll val, int lazy) : sum(sum), val(val), lazy(lazy) {}
49 void init() { sum = val = lazy = 0; }
50 node operator + (const node &other) const { return node(sum + other.sum, val + other.val, 0); }
51 void Xor() { val = sum - val; lazy ^= 1; }
52 }a[N << 2];
53 void build(int id, int l, int r)
54 {
55 a[id] = node(0, 0, 0);
56 if (l == r)
57 {
58 a[id].sum = by[l + 1] - by[l];
59 return;
60 }
61 int mid = (l + r) >> 1;
62 build(id << 1, l, mid);
63 build(id << 1 | 1, mid + 1, r);
64 a[id] = a[id << 1] + a[id << 1 | 1];
65 }
66 void pushdown(int id)
67 {
68 if (!a[id].lazy) return;
69 a[id << 1].Xor();
70 a[id << 1 | 1].Xor();
71 a[id].lazy = 0;
72 }
73 void update(int id, int l, int r, int ql, int qr)
74 {
75 if (l >= ql && r <= qr)
76 {
77 a[id].Xor();
78 return;
79 }
80 int mid = (l + r) >> 1;
81 pushdown(id);
82 if (ql <= mid) update(id << 1, l, mid, ql, qr);
83 if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr);
84 a[id] = a[id << 1] + a[id << 1 | 1];
85 }
86 }
87
88 int main()
89 {
90 while (scanf("%d", &n) != EOF)
91 {
92 mx = my = 0; res = 0;
93 for (int i = 1; i <= 2 * n; ++i) v[i].clear();
94 for (int i = 1; i <= n; ++i) rec[i].scan(); Hash();
95 for (int i = 1; i <= n; ++i)
96 {
97 v[rec[i].x[0]].emplace_back(rec[i].y[0], rec[i].y[1] - 1);
98 v[rec[i].x[1]].emplace_back(rec[i].y[0], rec[i].y[1] - 1);
99 }
100 n <<= 1;
101 SEG::build(1, 1, n);
102 for (int i = 1; i < mx; ++i)
103 {
104 for (auto it : v[i]) SEG::update(1, 1, n, it.first, it.second);
105 res += (bx[i + 1] - bx[i]) * SEG::a[1].val;
106 }
107 printf("%lld\n", res);
108 }
109 return 0;
110 }
G:Goat on a Rope
Solved.
签到。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 double x, y;
6 double xa, xb, ya, yb;
7
8 double calc(double xa, double ya, double xb, double yb)
9 {
10 return sqrt((xa - xb) * (xa - xb) + (ya - yb) * (ya - yb));
11 }
12
13 int main()
14 {
15 while(~scanf("%lf %lf %lf %lf %lf %lf", &x, &y, &xa, &ya, &xb, &yb))
16 {
17 double ans = 1e9;
18 if(x >= min(xa, xb) && x <= max(xa, xb)) ans = min(ans, min(fabs(y - ya), fabs(y - yb)));
19 if(y >= min(ya, yb) && y <= max(ya, yb)) ans = min(ans, min(fabs(x - xa), fabs(x - xb)));
20 ans = min(ans, calc(x, y, xa, ya));
21 ans = min(ans, calc(x, y, xa, yb));
22 ans = min(ans, calc(x, y, xb, ya));
23 ans = min(ans, calc(x, y, xb, yb));
24 printf("%.3f\n", ans);
25 }
26 return 0;
27 }
H:Repeating Goldbachs
Solved.
签到。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 1e6 + 10;
6
7 bool isprime[maxn];
8 int prime[maxn];
9
10 void Init()
11 {
12 memset(isprime, true, sizeof isprime);
13 isprime[0] = isprime[1] = false;
14 for(int i = 2; i < maxn; ++i)
15 {
16 if(isprime[i])
17 {
18 prime[++prime[0]] = i;
19 for(int j = i * 2; j < maxn; j += i)
20 {
21 isprime[j] = false;
22 }
23 }
24 }
25 }
26
27 int x;
28
29 int main()
30 {
31 Init();
32 while(~scanf("%d", &x))
33 {
34 int ans = 0;
35 while(x >= 4)
36 {
37 for(int i = 1; i <= prime[0]; ++i)
38 {
39 int tmp = prime[i];
40 if(isprime[x - tmp])
41 {
42 ++ans;
43 x = x - tmp - tmp;
44 break;
45 }
46 }
47 }
48 printf("%d\n", ans);
49 }
50 return 0;
51 }
I:Inversions
Unsolved.
题意：
给出一些序列，一些位置上的数字可以在$[1, k]的范围内随便填，求如果填使得整个序列的逆序对个数最多$
J:Time Limits
Solved.
签到。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 int n, s;
6
7 int main()
8 {
9 while(~scanf("%d %d", &n, &s))
10 {
11 int Max = -1;
12 for(int i = 1; i <= n; ++i)
13 {
14 int num;
15 scanf("%d", &num);
16 Max = max(Max, num);
17 }
18 Max *= s;
19 int ans = Max / 1000;
20 if(Max % 1000) ans++;
21 printf("%d\n", ans);
22 }
23 return 0;
24 }
K:Knockout
Unsolved.
题意：
给出一个数字，然后两个骰子的点数，每次可以移掉数字当中某几位加起来的和等于两骰子点数之和
那么就可以移掉这几位，知道最后不能移位置，最后剩下的数就是分数
现在给出一中间局面，求移掉哪些，使得最后的分数期望最大以及最小
L:Liars
Solved.
签到。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 1e3 + 10;
6
7 int n;
8 int cnt[maxn];
9
10 int main()
11 {
12 while(~scanf("%d", &n))
13 {
14 memset(cnt, 0, sizeof cnt);
15 for(int i = 1; i <= n; ++i)
16 {
17 int ai, bi;
18 scanf("%d %d", &ai, &bi);
19 for(int j = ai; j <= bi; ++j)
20 {
21 cnt[j]++;
22 }
23 }
24 int ans = -1;
25 for(int i = 1; i <= n; ++i)
26 {
27 if(cnt[i] == i) ans = i;
28 }
29 printf("%d\n", ans);
30 }
31 return 0;
32 }
M:Mobilization
Unsolved.
题意：
$有m种军队，每种军队有h_i 属性 和 p_i属性，以及购买一支军队需要c_i的钱，每种军队可以购买无限支$
$现在你有C个单位的钱，求如何购买军队，使得\sum h_i \cdot \sum p_i 最大$
原文地址：https://www.cnblogs.com/Dup4/p/10086500.html
时间： 2024-11-13 07:20:13