bupt summer training for 16 #3 ——构造

https://vjudge.net/contest/172464

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

 #include <cstdio>
 #include <vector>
 #include <algorithm>

 using std::vector;
 using std::sort;

 typedef long long ll;

 int n, m;

 ll a[], b[];

 ll sa, sb, dis, tmp, qaq;

 int t1 = -, t2 = -;

 int a1, a2, a3, a4;

 struct node {
     ll x, y, z;
     bool operator < (const node &a) const {
         return x < a.x;
     }
 };

 vector <node> e;

 ll abs(ll x) {
     return x <  ? -x : x;
 }

 node find1(ll x) {
     node ret = {, -, -};
     int l = , r = e.size() - , mid;
     while(l <= r) {
         int mid = (l + r) >> ;
         if(e[mid].x *  <= x) ret = e[mid], l = mid + ;
         else r = mid - ;
     }
     return ret;
 }

 node find2(ll x) {
     node ret = {, -, -};
     int l = , r = e.size() - , mid;
     while(l <= r) {
         int mid = (l + r) >> ;
         if(e[mid].x *  > x) ret = e[mid], r = mid - ;
         else l = mid + ;
     }
     return ret;
 }

 int main() {
     scanf("%d", &n);
     for(int i = ;i <= n;i ++)
         scanf("%lld", &a[i]), sa += a[i];
     scanf("%d", &m);
     for(int i = ;i <= m;i ++)
         scanf("%lld", &b[i]), sb += b[i];
     if(abs(sa - sb) <= ) {
         printf("%lld\n0\n", abs(sa - sb));
         return ;
     }
     qaq = tmp = dis = sa - sb;
     for(int i = ;i <= n;i ++)
         for(int j = ;j <= m;j ++)
             if(abs(qaq - (a[i] - b[j]) * ) < abs(dis))
                 dis = qaq - (a[i] - b[j]) * , t1 = i, t2 = j;
     for(int i = ;i <= n;i ++)
         for(int j = i + ;j <= n;j ++)
             e.push_back((node){a[i] + a[j], i, j});
     sort(e.begin(), e.end());
     for(int i = ;i <= m;i ++)
         for(int j = i + ;j <= m;j ++) {
             ll k = b[i] + b[j];
             node tt = find1(qaq + k * );
             if(tt.y != - && abs(qaq + k *  - tt.x * ) < abs(tmp)) tmp = qaq + k *  - tt.x * , a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
             tt = find2(qaq + k * );
             if(tt.y != - && abs(qaq + k *   - tt.x * ) < abs(tmp)) tmp = qaq + k *  - tt.x * , a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
         }
     if(abs(qaq) <= abs(dis) && abs(qaq) <= abs(tmp)) printf("%lld\n0\n", abs(qaq));
     else if(abs(dis) <= abs(tmp)) printf("%lld\n1\n%d %d", abs(dis), t1, t2);
     else printf("%lld\n2\n%d %d\n%d %d", abs(tmp), a1, a2, a3, a4);
     return ;
  }

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

 #include <cstdio>
 #include <vector>
 #include <algorithm>

 using std::vector;
 using std::sort;

 typedef long long ll;

 int n, m;

 ll a[], b[];

 ll sa, sb, dis, tmp, qaq;

 int t1 = -, t2 = -;

 int a1, a2, a3, a4;

 struct node {
     ll x, y, z;
     bool operator < (const node &a) const {
         return x < a.x;
     }
 };

 vector <node> e;

 ll abs(ll x) {
     return x <  ? -x : x;
 }

 node find1(ll x) {
     node ret = {, -, -};
     int l = , r = e.size() - , mid;
     while(l <= r) {
         int mid = (l + r) >> ;
         if(e[mid].x *  <= x) ret = e[mid], l = mid + ;
         else r = mid - ;
     }
     return ret;
 }

 node find2(ll x) {
     node ret = {, -, -};
     int l = , r = e.size() - , mid;
     while(l <= r) {
         int mid = (l + r) >> ;
         if(e[mid].x *  > x) ret = e[mid], r = mid - ;
         else l = mid + ;
     }
     return ret;
 }

 int main() {
     scanf("%d", &n);
     for(int i = ;i <= n;i ++)
         scanf("%lld", &a[i]), sa += a[i];
     scanf("%d", &m);
     for(int i = ;i <= m;i ++)
         scanf("%lld", &b[i]), sb += b[i];
     if(abs(sa - sb) <= ) {
         printf("%lld\n0\n", abs(sa - sb));
         return ;
     }
     qaq = tmp = dis = sa - sb;
     for(int i = ;i <= n;i ++)
         for(int j = ;j <= m;j ++)
             if(abs(qaq - (a[i] - b[j]) * ) < abs(dis))
                 dis = qaq - (a[i] - b[j]) * , t1 = i, t2 = j;
     for(int i = ;i <= n;i ++)
         for(int j = i + ;j <= n;j ++)
             e.push_back((node){a[i] + a[j], i, j});
     sort(e.begin(), e.end());
     for(int i = ;i <= m;i ++)
         for(int j = i + ;j <= m;j ++) {
             ll k = b[i] + b[j];
             node tt = find1(qaq + k * );
             if(tt.y != - && abs(qaq + k *  - tt.x * ) < abs(tmp)) tmp = qaq + k *  - tt.x * , a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
             tt = find2(qaq + k * );
             if(tt.y != - && abs(qaq + k *   - tt.x * ) < abs(tmp)) tmp = qaq + k *  - tt.x * , a1 = tt.y, a2 = i, a3 = tt.z, a4 = j;
         }
     if(abs(qaq) <= abs(dis) && abs(qaq) <= abs(tmp)) printf("%lld\n0\n", abs(qaq));
     else if(abs(dis) <= abs(tmp)) printf("%lld\n1\n%d %d", abs(dis), t1, t2);
     else printf("%lld\n2\n%d %d\n%d %d", abs(tmp), a1, a2, a3, a4);
     return ;
  }

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...

 #include <cmath>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 const double eps = 1e-;

 int n, L[], R[];

 double a[];

 bool dfs(int i, int x) {
     if(i > n)
         return ;
     int y = L[i] / x;
     if(L[i] % x) y ++;
     for(y *= x;y <= R[i];y += x)
         if(dfs(i + , y))
             return ;
     return ;
 }

 bool judge(double k) {
     for(int i = ;i <= n;i ++)
         L[i] = ceil(a[i] - a[i] * k), R[i] = floor(a[i] + a[i] * k);
     for(int i = L[];i <= R[];i ++)
         if(dfs(, i))
             return ;
     return ;
 }

 int main() {
     scanf("%d", &n);
     for(int i = ;i <= n;i ++)
         scanf("%lf", &a[i]);
     double l = , r = 1.0, mid, ans;
     for(int t = ;t --;) {
         mid = (l + r) / ;
         if(judge(mid)) ans = mid, r = mid - eps;
         else l = mid + eps;
     }
     printf("%.12f", ans);
     return ;
 }

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

 #include <cstdio>

 typedef long long ll;

 const int maxn = ;

 ll a[maxn], b[maxn];

 ll c(ll x) {
     return x * (x - ) * (x - ) / ;
 }

 int main() {
     for(int i = ;i <= ;i ++) a[i] = (i + ) /  - ;
     for(int i = , j = ;i <= ;i ++) {
         if(!(i & )) j ++;
         b[i] = b[i - ] + j;
     }
     for(int i = ;i <= ;i ++) a[i] += a[i - ], b[i] += b[i - ];
     int n;
     while(scanf("%d", &n), n >= ) printf("%lld\n", c(n) - a[n] - b[n]);
     return ;
 }

E.