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bupt summer training for 16 #3 ——构造

后来补题发现这场做的可真他妈傻逼

A.签到傻逼题，自己分情况

 1 #include <cstdio>
 2 #include <vector>
 3 #include <algorithm>
 4 
 5 using std::vector;
 6 using std::sort;
 7 
 8 typedef long long ll;
 9 
10 int n, m;
11 
12 ll a[2100], b[2100];
13 
14 ll sa, sb, dis, tmp, qaq;
15 
16 int t1 = -1, t2 = -1;
17 
18 int a1, a2, a3, a4;
19 
20 struct node {
21     ll x, y, z;
22     bool operator < (const node &a) const {
23         return x < a.x;
24     }
25 };
26 
27 vector <node> e;
28 
29 ll abs(ll x) {
30     return x < 0 ? -x : x;
31 }
32 
33 node find1(ll x) {
34     node ret = {0, -1, -1};
35     int l = 0, r = e.size() - 1, mid;
36     while(l <= r) {
37         int mid = (l + r) >> 1;
38         if(e[mid].x * 2 <= x) ret = e[mid], l = mid + 1;
39         else r = mid - 1;
40     }
41     return ret;
42 }
43 
44 node find2(ll x) {
45     node ret = {0, -1, -1};
46     int l = 0, r = e.size() - 1, mid;
47     while(l <= r) {
48         int mid = (l + r) >> 1;
49         if(e[mid].x * 2 > x) ret = e[mid], r = mid - 1;
50         else l = mid + 1;
51     }
52     return ret;
53 }
54 
55 int main() {
56     scanf("%d", &n);
57     for(int i = 1;i <= n;i ++)
58         scanf("%lld", &a[i]), sa += a[i];
59     scanf("%d", &m);
60     for(int i = 1;i <= m;i ++)
61         scanf("%lld", &b[i]), sb += b[i];
62     if(abs(sa - sb) <= 1) {
63         printf("%lld
0
", abs(sa - sb));
64         return 0;
65     }
66     qaq = tmp = dis = sa - sb;
67     for(int i = 1;i <= n;i ++) 
68         for(int j = 1;j <= m;j ++)
69             if(abs(qaq - (a[i] - b[j]) * 2) < abs(dis))
70                 dis = qaq - (a[i] - b[j]) * 2, t1 = i, t2 = j;
71     for(int i = 1;i <= n;i ++)
72         for(int j = i + 1;j <= n;j ++)
73             e.push_back((node){a[i] + a[j], i, j});
74     sort(e.begin(), e.end());
75     for(int i = 1;i <= m;i ++)
76         for(int j = i + 1;j <= m;j ++) {
77             ll k = b[i] + b[j];
78             node tt = find1(qaq + k * 2);
79             if(tt.y != -1 && abs(qaq + k * 2 - tt.x * 2) < abs(tmp)) tmp = qaq + k * 2 - tt.x * 2, a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
80             tt = find2(qaq + k * 2);
81             if(tt.y != -1 && abs(qaq + k *  2 - tt.x * 2) < abs(tmp)) tmp = qaq + k * 2 - tt.x * 2, a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
82         }
83     if(abs(qaq) <= abs(dis) && abs(qaq) <= abs(tmp)) printf("%lld
0
", abs(qaq));
84     else if(abs(dis) <= abs(tmp)) printf("%lld
1
%d %d", abs(dis), t1, t2);
85     else printf("%lld
2
%d %d
%d %d", abs(tmp), a1, a2, a3, a4);
86     return 0;
87  }

View Code

想写if-else结果写完if忘记else了

B.我是暴力的O(n^2logn)

首先如果只交换一个数，那O(n^2)都会算吧

那交换两个数呢，我把n个数两两结合并求和，再对他们排序

再枚举另一数组的数对，二分一下尝试更新答案

 1 #include <cstdio>
 2 #include <vector>
 3 #include <algorithm>
 4 
 5 using std::vector;
 6 using std::sort;
 7 
 8 typedef long long ll;
 9 
10 int n, m;
11 
12 ll a[2100], b[2100];
13 
14 ll sa, sb, dis, tmp, qaq;
15 
16 int t1 = -1, t2 = -1;
17 
18 int a1, a2, a3, a4;
19 
20 struct node {
21     ll x, y, z;
22     bool operator < (const node &a) const {
23         return x < a.x;
24     }
25 };
26 
27 vector <node> e;
28 
29 ll abs(ll x) {
30     return x < 0 ? -x : x;
31 }
32 
33 node find1(ll x) {
34     node ret = {0, -1, -1};
35     int l = 0, r = e.size() - 1, mid;
36     while(l <= r) {
37         int mid = (l + r) >> 1;
38         if(e[mid].x * 2 <= x) ret = e[mid], l = mid + 1;
39         else r = mid - 1;
40     }
41     return ret;
42 }
43 
44 node find2(ll x) {
45     node ret = {0, -1, -1};
46     int l = 0, r = e.size() - 1, mid;
47     while(l <= r) {
48         int mid = (l + r) >> 1;
49         if(e[mid].x * 2 > x) ret = e[mid], r = mid - 1;
50         else l = mid + 1;
51     }
52     return ret;
53 }
54 
55 int main() {
56     scanf("%d", &n);
57     for(int i = 1;i <= n;i ++)
58         scanf("%lld", &a[i]), sa += a[i];
59     scanf("%d", &m);
60     for(int i = 1;i <= m;i ++)
61         scanf("%lld", &b[i]), sb += b[i];
62     if(abs(sa - sb) <= 1) {
63         printf("%lld
0
", abs(sa - sb));
64         return 0;
65     }
66     qaq = tmp = dis = sa - sb;
67     for(int i = 1;i <= n;i ++) 
68         for(int j = 1;j <= m;j ++)
69             if(abs(qaq - (a[i] - b[j]) * 2) < abs(dis))
70                 dis = qaq - (a[i] - b[j]) * 2, t1 = i, t2 = j;
71     for(int i = 1;i <= n;i ++)
72         for(int j = i + 1;j <= n;j ++)
73             e.push_back((node){a[i] + a[j], i, j});
74     sort(e.begin(), e.end());
75     for(int i = 1;i <= m;i ++)
76         for(int j = i + 1;j <= m;j ++) {
77             ll k = b[i] + b[j];
78             node tt = find1(qaq + k * 2);
79             if(tt.y != -1 && abs(qaq + k * 2 - tt.x * 2) < abs(tmp)) tmp = qaq + k * 2 - tt.x * 2, a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
80             tt = find2(qaq + k * 2);
81             if(tt.y != -1 && abs(qaq + k *  2 - tt.x * 2) < abs(tmp)) tmp = qaq + k * 2 - tt.x * 2, a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
82         }
83     if(abs(qaq) <= abs(dis) && abs(qaq) <= abs(tmp)) printf("%lld
0
", abs(qaq));
84     else if(abs(dis) <= abs(tmp)) printf("%lld
1
%d %d", abs(dis), t1, t2);
85     else printf("%lld
2
%d %d
%d %d", abs(tmp), a1, a2, a3, a4);
86     return 0;
87  }

View Code

补题的时候，就把比赛代码三个地方的 int 改成了long long就过了

C.别人补题写的DP一眼看不懂...反正数据范围也不大，我们来xjb乱搞吧

数据范围20，时间5s，没有直接爆搜的思路

答案在0-1之间，满足单调性...试试二分？那judge呢？暴力dfs枚举?

效率玄学？并没有关系！...就当试试咯...过了...比DP还快...

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const double eps = 1e-14;
 8 
 9 int n, L[21], R[21];
10 
11 double a[21];
12 
13 bool dfs(int i, int x) {
14     if(i > n)
15         return 1;
16     int y = L[i] / x;
17     if(L[i] % x) y ++;
18     for(y *= x;y <= R[i];y += x)
19         if(dfs(i + 1, y))
20             return 1;
21     return 0;
22 }
23 
24 bool judge(double k) {
25     for(int i = 1;i <= n;i ++)
26         L[i] = ceil(a[i] - a[i] * k), R[i] = floor(a[i] + a[i] * k);
27     for(int i = L[1];i <= R[1];i ++)
28         if(dfs(2, i))
29             return 1;
30     return 0;
31 }
32 
33 int main() {
34     scanf("%d", &n);
35     for(int i = 1;i <= n;i ++)
36         scanf("%lf", &a[i]);
37     double l = 0, r = 1.0, mid, ans;
38     for(int t = 66;t --;) {
39         mid = (l + r) / 2;
40         if(judge(mid)) ans = mid, r = mid - eps;
41         else l = mid + eps;
42     }
43     printf("%.12f", ans);
44     return 0;
45 }

View Code

看了别人DP代码...看懂了...

f[i][j]代表把第 i 种货币变成 j 的最小答案

我们有一种无赖解是把所有货币都变0

所以对于第 i 种货币，从 0 枚举到 a[i] * 2 就可以啦

复杂度O(nklogk)， k = max(a[i]) = 20W

这么来说粗略计算我的做法复杂度就是O(nklogk * log(1/eps))...实际优太多

D.ans = C(n，3) - 不合法的三角形

对于非法三角形枚举最大边 z

再求 x + y <= z 的 (x，y) 有多少对即可

预处理，O(1)回答

 1 #include <cstdio>
 2 
 3 typedef long long ll;
 4 
 5 const int maxn = 1000010;
 6 
 7 ll a[maxn], b[maxn];
 8 
 9 ll c(ll x) {
10     return x * (x - 1) * (x - 2) / 6;
11 }
12 
13 int main() {
14     for(int i = 3;i <= 1000000;i ++) a[i] = (i + 1) / 2 - 1;
15     for(int i = 4, j = 0;i <= 1000000;i ++) {
16         if(!(i & 1)) j ++;
17         b[i] = b[i - 1] + j;
18     }    
19     for(int i = 3;i <= 1000000;i ++) a[i] += a[i - 1], b[i] += b[i - 1];
20     int n;
21     while(scanf("%d", &n), n >= 3) printf("%lld
", c(n) - a[n] - b[n]);
22     return 0;
23 }

View Code

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原文地址：https://www.cnblogs.com/ytytzzz/p/7253450.html