Codeforces Round #558 (div. 2)
作者：互联网
bc两题居然都有小题，有点烦。
题目链接：http://codeforces.com/contest/1163
A：
本场唯一温暖人心。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define int_inf 0x3f3f3f3f
33 #define ll_inf 0x7f7f7f7f7f7f7f7f
34 #define lson curPos<<1
35 #define rson curPos<<1|1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 int n, m;
41
42 int main()
43 {
44 scanf("%d%d", &n, &m);
45 if (!m) puts("1");
46 else
47 {
48 if (m <= n / 2) printf("%d\n", m);
49 else printf("%d\n", n - m);
50 }
51 return 0;
52 }
View Code
B easy & hard：
读懂题之后其实也挺傻的。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define int_inf 0x3f3f3f3f
33 #define ll_inf 0x7f7f7f7f7f7f7f7f
34 #define lson curPos<<1
35 #define rson curPos<<1|1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 const int maxn = 1e5 + 10;
41 int n, a[maxn], cnt[maxn], b[maxn], used, ans = 1;
42
43 int main()
44 {
45 scanf("%d", &n);
46 rep1(i, 1, n) scanf("%d", &a[i]);
47 rep1(i, 1, n)
48 {
49 if (!cnt[a[i]]) used++;
50 b[cnt[a[i]]]--; cnt[a[i]]++; b[cnt[a[i]]]++;
51 if (b[cnt[a[i]]] == used - 1 && b[cnt[a[i]] + 1] == 1) ans = i;
52 if (b[cnt[a[i]]] == used - 1 && b[1] == 1) ans = i;
53 if (b[cnt[a[i]]] == 1 && b[cnt[a[i]] - 1] == used - 1) ans = i;
54 if (b[cnt[a[i]]] == used && i != n) ans = i + 1;
55 }
56 printf("%d\n", ans);
57 return 0;
58 }
View Code
C easy & hard：
easy版本暴力就可以了，hard版本稍难一些。从n<=1000就可以猜到大概是个O(n^2)的做法。
1 /* basic header */
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <string>
6 #include <cstring>
7 #include <cmath>
8 #include <cstdint>
9 #include <climits>
10 #include <float.h>
11 /* STL */
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <queue>
16 #include <stack>
17 #include <algorithm>
18 #include <array>
19 #include <iterator>
20 /* define */
21 #define ll long long
22 #define dou double
23 #define pb emplace_back
24 #define mp make_pair
25 #define fir first
26 #define sec second
27 #define sot(a,b) sort(a+1,a+1+b)
28 #define rep1(i,a,b) for(int i=a;i<=b;++i)
29 #define rep0(i,a,b) for(int i=a;i<b;++i)
30 #define repa(i,a) for(auto &i:a)
31 #define eps 1e-8
32 #define inf 0x3f3f3f3f
33 #define ll_inf 0x7f7f7f7f7f7f7f7f
34 #define lson curPos<<1
35 #define rson curPos<<1|1
36 /* namespace */
37 using namespace std;
38 /* header end */
39
40 const int maxn = 1e3 + 10;
41 struct Point
42 {
43 int x, y;
44 };
45
46 map<pair<pair<ll, ll>, pair<ll, ll> >, ll>m;
47 map<pair<ll, ll>, ll>mm;
48 Point p[maxn];
49 int n;
50 ll ans = 0;
51
52 int main()
53 {
54 m.clear(); mm.clear();
55 scanf("%d", &n);
56 for (int i = 1; i <= n; i++)
57 {
58 scanf("%lld%lld", &p[i].x, &p[i].y);
59 for (int j = 1; j < i; j++)
60 {
61 int xx = p[i].x - p[j].x, yy = p[i].y - p[j].y, zz = xx * p[j].y - yy * p[j].x;
62 int gcd = __gcd(xx, yy), gcd2 = __gcd(xx, zz);
63 if (!xx)
64 m[ {{xx / gcd, yy / gcd}, {p[i].x, inf}}]++;
65 else
66 m[ {{xx / gcd, yy / gcd}, {xx / gcd2, zz / gcd2}}]++;
67 }
68 }
69 for (auto i : m)
70 mm[ {i.fir.fir.fir, i.fir.fir.sec}]++;
71 for (auto id : m)
72 ans += (int)m.size() - mm[ {id.fir.fir.fir, id.fir.fir.sec}];
73 printf("%lld\n", ans / 2);
74 return 0;
75 }
View Code
D：
看了题目莫得想法，往右下角一看tag是dp string，打扰了。
标签：558,int,ll,Codeforces,long,pair,div,include,define 来源： https://www.cnblogs.com/JHSeng/p/10845164.html