[Gym]2008-2009 ACM-ICPC, NEERC, Moscow Subregional Contest

比赛链接：http://codeforces.com/gym/100861

A模拟，注意两个特殊的缩写。

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

 typedef pair<string, string> pss;
 const string MSU = "MSU";
 const string SCH = "SCH";
 const int maxn = ;
 map<string, int> school;
 int n;
 string a, b;
 vector<pss> team;
 vector<pss> ret;

 string up(string x) {
   int len = x.length();
   for(int i = ; i < len; i++) {
     if(x[i] >= 'a' && x[i] <= 'z') x[i] = x[i] - 'a' + 'A';
   }
   return x;
 }

 int main() {
   //freopen("in", "r", stdin);
   while(~scanf("%d", &n)) {
     school.clear(); team.clear(); ret.clear();
     for(int i = ; i <= n; i++) {
       cin >> a >> b;
       team.push_back(pss(a, b));
     }
     for(int i = ; i < n; i++) {
       string tmp = up(team[i].first);
       if(team[i].first == SCH) continue;
       if(school.find(tmp) == school.end()) {
         ret.push_back(team[i]);
         school[tmp] = ;
         continue;
       }
       if(team[i].first == MSU) {
         if(school[tmp] < ) {
           ret.push_back(team[i]);
           school[tmp]++;
         }
       }
       else {
         if(school[tmp] < ) {
           ret.push_back(team[i]);
           school[tmp]++;
         }
       }
     }
     if(ret.size() <= ) {
       printf("%d\n", ret.size());
       for(int i = ; i < ret.size(); i++) {
         cout << ret[i].first <<" " << ret[i].second << endl;
       }
     }
     else {
       printf("%d\n", );
       for(int i = ; i < ; i++) {
         cout << ret[i].first <<" " << ret[i].second << endl;
       }
     }
   }
   return ;
 }

B排序+离散化

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

 typedef long long LL;
 typedef struct F {
   int idx;
   LL val;
 }F;
 typedef struct S {
   int idx;
   int val;
 }S;
 const int maxn = ;
 const int maxm = ;
 S s[maxm];
 int k;
 int n, m;
 LL a, b, c;
 F f[maxm];
 LL ret[maxm];

 int h[maxm], hcnt;
 int t[maxm], tcnt;

 bool cmp1(S a, S b) {
   if(a.val == b.val) return a.idx < b.idx;
   return a.val > b.val;
 }

 bool cmp2(F a, F b) {
   if(a.val == b.val) return a.idx < b.idx;
   return a.val > b.val;
 }

 int id(int x) {
   return lower_bound(h, h+hcnt, x) - h + ;
 }

 int main() {
   freopen("in", "r", stdin);
   int x;
   while(~scanf("%I64d%I64d%I64d%I64d",&f[].val,&a,&b,&c)) {
     memset(ret, , sizeof(ret));
     k = ; hcnt = , tcnt = ;
     for(int i = ; i < maxm; i++) {
       s[i].idx = -;
       s[i].val = ;
     }
     scanf("%d%d",&n,&m);
     for(int i = ; i <= n; i++) {
       for(int j = ; j <= m; j++) {
         scanf("%d", &x);
         t[tcnt++] = x;
         h[hcnt++] = x;
       }
     }
     sort(h, h+hcnt); hcnt = unique(h, h+hcnt) - h;
     for(int i = ; i < tcnt; i++) {
       if(s[id(t[i])].idx == -) {
         k++;
         s[id(t[i])].idx = t[i];
       }
       s[id(t[i])].val++;
     }
     sort(s+, s+maxm, cmp1);
     f[].idx = ;
     for(int i = ; i <= k; i++) {
       f[i].idx = i;
       f[i].val = ((a * f[i-].val + b) % c) + ;
     }
     sort(f+, f+k+, cmp2);
     for(int i = ; i<= k; i++) {
       ret[f[i].idx] = s[i].idx;
     }
     printf("%d\n", k);
     for(int i = ; i<= k; i++) {
       printf("%I64d%c", ret[i], i == k ? '\n' : ' ');
     }
   }
   return ;
 }

C贪心，就像搭积木一样，全是1*1*1的小方块，要想让表面积尽可能小，那就把他们往一个角落堆。比如

3 3

1 2 3 4 5 6 7 8 9这样的数据，那么排出来就是。

9 8 5

7 6 4

3 2 1

这样可以保证遮盖的部分最大。那么只需要找到第一行和第一列的高度，两边加起来乘2就行了。

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

 const int maxn = ;
 int n, m, k;
 int t[maxn*maxn];

 int main() {
   // freopen("in", "r", stdin);
   while(~scanf("%d%d",&m,&n)) {
     k = ;
     if(m < n) swap(m, n);
     for(int i = ; i <= m; i++) {
       for(int j = ; j <= n; j++) {
         scanf("%d",&t[k++]);
       }
     }
     sort(t, t+k, greater<int>());
     int ret = ;
     for(int i = ; i < n; i++) {
       ret += t[i*i];
       ret += t[i*(i+)];
     }
     for(int i = n; i < m; i++) {
       ret += t[i*n];
     }
     cout << ret *  << endl;
   }
   return ;
 }

G首先知道结果为0的情况，那就是小于等于3个点的时候；想要结果为1，那么必然是两两配对出来的新点位置重合。否则可以让任意四个点组成一个四边形，这个时候分别取每边的中点，构成的四边形一定是平行四边形（很好证明）。所以第二次就一定会有交点了（对角线中心）。

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

 typedef pair<int,int> pii;
 const int maxn = ;
 int n;
 int x[maxn], y[maxn];
 map<pii, bool> s;

 int main() {
   // freopen("in", "r", stdin);
   while(~scanf("%d", &n)) {
     s.clear();
     for(int i = ; i <= n; i++) {
       scanf("%d%d",&x[i],&y[i]);
       x[i] <<= ; y[i] <<= ;
     }
     if(n <= ) {
       puts("");
       continue;
     }
     bool flag = ;
     for(int i = ; i <= n; i++) {
       if(flag) break;
       for(int j = i+; j <= n; j++) {
         int tx = (x[i] + x[j]) / ;
         int ty = (y[i] + y[j]) / ;
         if(s.find(pii(tx, ty)) != s.end()) {
           printf("1\n");
           flag = ;
           break;
         }
         s[pii(tx, ty)] = ;
       }
     }
     if(!flag) puts("");
   }
   return ;
 }

L打表找规律，然后就很简单了。

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

 const int maxn = ;
 bool vis[maxn];

 bool lucky(int s) {
   int x, y;
   x = s % ;
   s /= ;
   y = s;
   int sx = , sy = ;
   while(x) {
     sx += x % ;
     x /= ;
   }
   while(y) {
     sy += y % ;
     y /= ;
   }
   return sx == sy;
 }

 int main() {
   memset(vis, , sizeof(vis));
   for(int i = ; i <= ; i++) {
     vis[i] = lucky(i);
   }
   int ret = ;
   int tmp = ;
   int rl ,rh;
   int lo;
   lo = ;
   for(int i = ; i <= ; i++) {
     if(vis[i]) {
       if(ret < tmp) {
         rl = lo;
         rh = i - ;
         ret = tmp;
       }
       tmp = ;
       lo = i + ;
     }
     else {
       tmp++;
     }
   }
   cout <<ret <<endl;
   cout << rl << " " << rh << endl;
 }

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

 string ret[][] = {
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
   {"",""},
 };
 int main() {
   int n;
   while(~scanf("%d", &n)) {
     cout << ret[n][] << endl << ret[n][] << endl;
   }
 }