https://vjudge.net/contest/172464

后來補題發現這場做的可真他媽傻逼

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A.簽到傻逼題，自己分情況

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\n0\n", 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\n0\n", abs(qaq)); 84 else if(abs(dis) <= abs(tmp)) printf("%lld\n1\n%d %d", abs(dis), t1, t2); 85 else printf("%lld\n2\n%d %d\n%d %d", abs(tmp), a1, a2, a3, a4); 86 return 0; 87 } View Code

想寫if-else結果寫完if忘記else了

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B.我是暴力的O(n^2logn)

首先如果只交換一個數，那O(n^2)都會算吧

那交換兩個數呢，我把n個數兩兩結合并求和，再對他們排序

再枚舉另一數組的數對，二分一下嘗試更新答案

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\n0\n", 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\n0\n", abs(qaq)); 84 else if(abs(dis) <= abs(tmp)) printf("%lld\n1\n%d %d", abs(dis), t1, t2); 85 else printf("%lld\n2\n%d %d\n%d %d", abs(tmp), a1, a2, a3, a4); 86 return 0; 87 } View Code

補題的時候，就把比賽代碼三個地方的 int 改成了long long就過了

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C.別人補題寫的DP一眼看不懂...反正數據范圍也不大，我們來xjb亂搞吧

數據范圍20，時間5s，沒有直接爆搜的思路

答案在0-1之間，滿足單調性...試試二分？那judge呢？暴力dfs枚舉?

效率玄學？并沒有關系！...就當試試咯...過了...比DP還快...

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

看了別人DP代碼...看懂了...

f[i][j]代表把第 i 種貨幣變成 j 的最小答案

我們有一種無賴解是把所有貨幣都變0

所以對于第 i 種貨幣，從 0 枚舉到 a[i] * 2 就可以啦

復雜度O(nklogk)， ?k = max(a[i]) = 20W

這么來說粗略計算我的做法復雜度就是O(nklogk * log(1/eps))...實際優太多

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D.ans = C(n，3) - 不合法的三角形

對于非法三角形枚舉最大邊 z

再求 x + y <= z 的 (x，y) 有多少對即可

預處理，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\n", c(n) - a[n] - b[n]); 22 return 0; 23 } View Code

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E.

轉載于:https://www.cnblogs.com/ytytzzz/p/7253450.html

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