CODEFORCES ROUND #761 ANALYSES BY TEAM:RED & BLACK

A. Dasha and Stairs

Problems: 一个按照1,2,3……编号的楼梯，给定踩过的编号为奇数奇数和偶数的楼梯数量a和b，问是否可以有区间[l, r]符合奇数编号有a个，偶数编号有b个。

Analysis:

　　cj: 纸张的我=.= 经过Return改正，才发现没有主义a = b = 0的情况。

 #define PRON "a"
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 #define inf 0x3f3f3f3f
 #define LL "%lld"
 using namespace std;
 typedef long long ll;

 int a, b;

 int main(){
     scanf("%d %d", &a, &b);
     if ((!a) && (!b)) { puts("NO"); return ; }
     if (abs(a - b) <= )
         cout << "YES";
     else
         cout << "NO";
 }

Code by Return

B. Dasha and friends

Problems:有无数不同的长度为l的环，上面有m个障碍。有两个人在不同的位置，给出障碍相对于自己的距离。判断两人是否在同一个环上。

Analysis:

　　cj: 做个差，随便比较一下。

 #define PRON "b"
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 #define inf 0x3f3f3f3f
 #define LL "%lld"
 using namespace std;
 typedef long long ll;

 const int maxn =  + ;

 int n, l, a[maxn], b[maxn], c[maxn], d[maxn];

 bool ok(int x){
     for (int i = , j = x; i < n; i ++, j ++){
         if (j == n)
             j = ;
         if (c[i] != d[j])
             return false;
     }

     return true;
 }

 bool check(){
     for (int i = ; i < n; i ++)
             if (c[] == d[i] && ok(i))
                 return true;
     return false;
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen(PRON ".in", "r", stdin);
 #endif

     cin >> n >> l;
     for (int i = ; i < n; i ++){
         cin >> a[i];
         if (i)
             c[i] = a[i] - a[i - ];
     }
     c[] = a[] + l - a[n - ];
     for (int i = ; i < n; i ++){
         cin >> b[i];
         if (i)
             d[i] = b[i] - b[i - ];
     }
     d[] = b[] + l - b[n - ];

     if (check())
         cout << "YES";
     else
         cout << "NO";
 }

Code by cj

C. Dasha and Password

Problems:一个合格的密码需要有至少一个字母、至少一个数字、至少一个特殊字符（* # &）。有n个字符串，光标都在第一个字母处。（光标在最左向右移动就到最右）求最少的光标移动操作，使得光标所选中的字母能构成一个合格的密码。

Analysis:

　　cj: 对于每个字符串求出最近的字母、数字、特殊字符的最小操作。数据范围小，直接for一下就好。

 #define PRON "c"
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 #define inf 0x3f3f3f3f
 #define LL "%lld"
 #define ed second
 #define st first
 using namespace std;
 typedef long long ll;

 const int maxn =   + ;

 int n, len;
 pair<int, int> dig[maxn], sy[maxn], l[maxn];
 string s;

 int get_ans(){
     int ans = inf;
     for (int i = ; i < n; i ++)
         for (int j = ; j < n; j ++)
             for (int k = ; k < n; k ++){
                 if (sy[i].ed != l[j].ed && l[j].ed != dig[k].ed && dig[k].ed != sy[i].ed){
                     if (~sy[i].st && ~l[j].st && ~dig[k].st)
                         ans = min(sy[i].st + l[j].st + dig[k].st, ans);
                 }
             }
     return ans;
 }

 inline bool is_dig(char c){
     return '' <= c && c <= '';
 }

 inline bool is_sy(char c){
     return c == '#' || c == '*' || c == '&';
 }

 inline bool is_l(char c){
     return ('a' <= c && c <= 'z') || ('A' <= c && c <= 'Z');
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen(PRON ".in", "r", stdin);
 #endif

     cin >> n >> len;
     for (int i = ; i < n; i ++){
         cin >> s;

         s[len] = '\0';
         dig[i].st = sy[i].st = l[i].st = -;
         dig[i].ed = sy[i].ed = l[i].ed = i;
         for (int j = , k = len; j <= k; j ++, k --){
             if (l[i].st == - && (is_l(s[j]) || is_l(s[k])))
                 l[i].st = j;
             if (dig[i].st == -  && (is_dig(s[j]) || is_dig(s[k])))
                 dig[i].st = j;
             if (sy[i].st == -  && (is_sy(s[j]) || is_sy(s[k])))
                 sy[i].st = j;

             if (~dig[i].st && ~sy[i].st && ~l[i].st)
                 break;
         }
     }

     sort(l, l + n);
     sort(dig, dig + n);
     sort(sy, sy + n);

     cout << get_ans();
 }

Code by cj

D. Dasha and Very Difficult Problem

Problems: c[i] = b[i] - a[i]，给定l, r, n、数组a、数组c的大小排名，求符合条件的数组b。其中l <= a[i], b[i] <= r。

Analysis:

　　return:

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 //#include<iostream>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<string>
 #include<vector>
 using namespace std;
 #define INF 2000000000
 #define Clear(x, Num) memset(x, Num, sizeof(x))
 #define Dig(x) ((x>='0') && (x<='9'))
 #define Neg(x) (x=='-')
 #define G_c() getchar()
 #define Maxn 1000010
 typedef long long ll;

 struct Data
 {
     int rank, id;
     ll l, r;
 }c[Maxn];
 int n;
 ll l, r, a[Maxn], b[Maxn];

 inline int gcd(int x, int y) { if (!y) return x; return gcd(y, x%y); }
 inline void read(int &x){ char ch; int N=; while ((ch=G_c()) && (!Dig(ch)) && (!Neg(ch))); if (Neg(ch)) { N=-; while ((ch=G_c()) && (!Dig(ch))); } x=ch-; while ((ch=G_c()) && (Dig(ch))) x=x*+ch-; x*=N; }
 //inline void Insert(int u, int v) { To[Cnt]=v; Next[Cnt]=Head[u]; Head[u]=Cnt++; }
 inline bool cmp(const Data &a, const Data &b) { return a.rank<b.rank; }
 int main()
 {
     scanf("%d%lld%lld", &n, &l, &r);
     for (int i=; i<=n; i++) scanf("%lld", &a[i]);
     for (int i=; i<=n; i++) scanf("%d", &c[i].rank), c[i].id=i;
     sort(c+, c+n+, cmp);
     c[].l=l-a[c[].id]; c[].r=r-a[c[].id];
     for (int i=; i<=n; i++)
     {
         c[i].l=max(l-a[c[i].id], c[i-].l+);
         c[i].r=min(r-a[c[i].id], c[i-].r+);
         if (c[i].l>c[i].r) { puts("-1"); return ; }
     }
     b[c[n].id]=c[n].l;
     for (int i=n-; i>=; i--)
     {
         c[i].r=min(c[i].r, b[c[i+].id]);
         if (c[i].l>c[i].r) { puts("-1"); return ; }
         b[c[i].id]=c[i].l;
     }
     for (int i=; i<=n; i++) printf("%lld ", b[i]+a[i]); puts("");
 }

code by return

E. Dasha and Puzzle

Problems: 给一颗无根数，把它画在平面上。边长随意，边必须和坐标轴平行，边之间不能在非端点处相交。

Analysis:

　　cj: 边长随意！！！n <= 30, |x|, |y| <= 1e18！！！把1当作根，dep=1时把len弄个2的很多次方，深度每+1，len就/2，这样每个点就随意能的有三个儿子。

 #define PRON "e"
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 #define inf 0x3f3f3f3f
 #define LL "%lld"
 #define st first
 #define nd second
 using namespace std;
 typedef long long ll;

 #define mp make_pair

 const int maxn =  + ;
 const int ax[] = {-, , , };
 const int ay[] = {, , , -};

 vector<int> g[maxn];
 pair<ll, ll> ans[maxn];
 int n;

 inline void fail(){
     printf("NO");
     exit();
 }

 void dfs(int u, int pre, ll len, int dir, ll x, ll y){
     if (g[u].size() > )
         fail();

     ans[u] = mp(x, y);
     for (int i = , v, now_dir = ; i < g[u].size(); i ++){
         v = g[u][i];
         if (v == pre)
             continue;

         if (pre !=  && (dir + ) %  == now_dir)
             now_dir ++;

         dfs(v, u, len / , now_dir, x + len * (ll)ax[now_dir], y + len * (ll)ay[now_dir]);
         now_dir ++;
     }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen(PRON ".in", "r", stdin);
 #endif

     cin >> n;
     for (int i = , u, v; i < n - ; i ++){
         cin >> u >> v;
         g[u].push_back(v);
         g[v].push_back(u);
     }

     dfs(, , 1ll << , , 0ll, 0ll);

     cout << "YES\n";
     for (int i = ; i <= n; i ++)
         cout << ans[i].st << " " << ans[i].nd << endl;
 }

code by cj

F. Dasha and Photos

Problems:

Analysis: