hihoCoder太阁最新面经算法竞赛17

比赛链接：http://hihocoder.com/contest/hihointerview26

A.排序后枚举两个点，确定一个矩形后二分剩下两个点。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 typedef struct Point {
     int x, y;
     Point() {}
     Point(int xx, int yy) : x(xx), y(yy) {}
 }Point;
 int n;
 LL ans;
 Point p[maxn];

 bool cmp(Point a, Point b) {
     if(a.x == b.x) return a.y < b.y;
     return a.x < b.x;
 }

 bool bs(int x, int y) {
     int ll = , rr = n, mm;
     while(ll <= rr) {
         mm = (ll + rr) >> ;
         if(p[mm].x == x && p[mm].y == y) return ;
         else if(cmp(p[mm], Point(x, y))) ll = mm + ;
         else rr = mm - ;

     }
     return ;
 }

 int main() {
     // freopen("in", "r", stdin);
     int x3, y3, x4, y4;
     while(~scanf("%d", &n) && n) {
         ans = 1000000LL * 1001000LL;
         for(int i = ; i < n; i++) {
             scanf("%d%d", &p[i].x, &p[i].y);
         }
         sort(p, p+n, cmp);
         for(int i = ; i < n; i++) {
             for(int j = i + ; j < n; j++) {
                 x3 = p[i].x; y3 = p[j].y;
                 x4 = p[j].x; y4 = p[i].y;
                 if(bs(x3, y3) && bs(x4, y4)) {
                     int a = x3 - x4;
                     int b = y3 - y4;
                     if((LL)a * b == ) continue;
                     ans = min(ans, abs((LL)a*b));
                 }
             }
         }
         printf("%lld\n", ans == 1000000LL * 1001000LL ? - : ans);
     }
     return ;
 }

B.按题要求爆搜

 #include <bits/stdc++.h>
 using namespace std;

 const int maxn = ;
 int n;
 set<string> ret;
 set<string>::iterator it;
 int num[maxn];
 set<int> pos;

 int main() {
     // freopen("in", "r", stdin);
     // freopen("out", "w", stdout);
     for(int i = ; i < maxn; i++) num[i] = i;
     while(~scanf("%d", &n)) {
         ret.clear();
         do {
             int nn = ( << (n));
             for(int i = ; i < nn; i++) {
                 pos.clear();
                 int tmp = i;
                 int cnt = ;
                 while(tmp) {
                     if(tmp & ) pos.insert(cnt+);
                     tmp >>= ; cnt++;
                 }
                 string t = "";
                 bool ex = ;
                 t += (num[] + '');
                 int pre = num[];
                 for(int i = ; i <= n; i++) {
                     if(ex) break;
                     if(pos.find(i) != pos.end()) {
                         t += '-';
                         pre = -;
                     }
                     if(pre > num[i]) {
                         ex = ;
                         break;
                     }
                     t += (num[i] + '');
                     pre = num[i];
                 }
                 if(!ex) ret.insert(t);
             }
             // for(int i = 1; i <= n; i++) printf("%d", num[i]);
             // printf("\n");
         }while(next_permutation(num+, num+n+));
         // printf("%d\n", ret.size());
         for(it = ret.begin(); it != ret.end(); it++) {
             cout << *it << endl;
         }
     }
     return ;
 }

C.找到规律后斯特灵数胡搞（好像没必要？）

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 const LL mod = 1000000007LL;
 LL f[maxn], a[maxn];
 LL S[maxn][maxn];
 int n;

 LL exgcd(LL a, LL b, LL &x, LL &y) {
     if(b == ) {
         x = ;
         y = ;
         return a;
     }
     else {
         LL ret = exgcd(b, a%b, x, y);
         LL tmp = x;
         x = y;
         y = tmp - a / b * y;
         return ret;
     }
 }

 LL inv(LL a) {
     LL x, y;
     exgcd(a, mod, x, y);
     return (x % mod + mod) % mod;
 }

 LL C(LL n, LL m) {
     return f[n] * inv(f[m]) % mod * inv(f[n-m]) % mod;
 }

 LL mul(LL x, LL n) {
     LL ret = ;
     while(n) {
         if(n & ) ret = ret * x % mod;
         n >>= ;
         x = x * x % mod;
     }
     return ret;
 }

 int main() {
     // freopen("in", "r", stdin);
     // freopen("out", "w", stdout);
     memset(S, , sizeof(S));
     S[][] = ;
     for(int p = ; p < maxn; p++) {
         for(int k = ; k < maxn; k++) {
             S[p][k] = (LL)k * S[p-][k] % mod + S[p-][k-] % mod;
         }
     }
     f[] = ;
     for(int i = ; i < maxn; i++) {
         f[i] = f[i-] * (LL)i % mod;
     }
     memset(a, , sizeof(a));
     a[] = ; a[] = ;
     for(int i = ; i <= ; i++) {
         for(int k = ; k <= i; k++) {
             a[i] = (a[i] + f[k] * S[i][k] % mod) % mod;
         }
     }
     while(~scanf("%d", &n)) {
         printf("%lld\n", a[n]);
     }
     return ;
 }