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。
1 #define PRON "a" 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 #define inf 0x3f3f3f3f 8 #define LL "%lld" 9 using namespace std; 10 typedef long long ll; 11 12 int a, b; 13 14 int main(){ 15 scanf("%d %d", &a, &b); 16 if ((!a) && (!b)) { puts("NO"); return 0; } 17 if (abs(a - b) <= 1) 18 cout << "YES"; 19 else 20 cout << "NO"; 21 }
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。
1 #define PRON "b" 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 #define inf 0x3f3f3f3f 8 #define LL "%lld" 9 using namespace std; 10 typedef long long ll; 11 12 const int maxn = 100 + 10; 13 14 int n, l, a[maxn], b[maxn], c[maxn], d[maxn]; 15 16 bool ok(int x){ 17 for (int i = 0, j = x; i < n; i ++, j ++){ 18 if (j == n) 19 j = 0; 20 if (c[i] != d[j]) 21 return false; 22 } 23 24 return true; 25 } 26 27 bool check(){ 28 for (int i = 0; i < n; i ++) 29 if (c[0] == d[i] && ok(i)) 30 return true; 31 return false; 32 } 33 34 int main(){ 35 #ifndef ONLINE_JUDGE 36 freopen(PRON ".in", "r", stdin); 37 #endif 38 39 cin >> n >> l; 40 for (int i = 0; i < n; i ++){ 41 cin >> a[i]; 42 if (i) 43 c[i] = a[i] - a[i - 1]; 44 } 45 c[0] = a[0] + l - a[n - 1]; 46 for (int i = 0; i < n; i ++){ 47 cin >> b[i]; 48 if (i) 49 d[i] = b[i] - b[i - 1]; 50 } 51 d[0] = b[0] + l - b[n - 1]; 52 53 if (check()) 54 cout << "YES"; 55 else 56 cout << "NO"; 57 }
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。
1 #define PRON "c" 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 #define inf 0x3f3f3f3f 8 #define LL "%lld" 9 #define ed second 10 #define st first 11 using namespace std; 12 typedef long long ll; 13 14 const int maxn = 50 + 10; 15 16 int n, len; 17 pair<int, int> dig[maxn], sy[maxn], l[maxn]; 18 string s; 19 20 int get_ans(){ 21 int ans = inf; 22 for (int i = 0; i < n; i ++) 23 for (int j = 0; j < n; j ++) 24 for (int k = 0; k < n; k ++){ 25 if (sy[i].ed != l[j].ed && l[j].ed != dig[k].ed && dig[k].ed != sy[i].ed){ 26 if (~sy[i].st && ~l[j].st && ~dig[k].st) 27 ans = min(sy[i].st + l[j].st + dig[k].st, ans); 28 } 29 } 30 return ans; 31 } 32 33 inline bool is_dig(char c){ 34 return '0' <= c && c <= '9'; 35 } 36 37 inline bool is_sy(char c){ 38 return c == '#' || c == '*' || c == '&'; 39 } 40 41 inline bool is_l(char c){ 42 return ('a' <= c && c <= 'z') || ('A' <= c && c <= 'Z'); 43 } 44 45 int main(){ 46 #ifndef ONLINE_JUDGE 47 freopen(PRON ".in", "r", stdin); 48 #endif 49 50 cin >> n >> len; 51 for (int i = 0; i < n; i ++){ 52 cin >> s; 53 54 s[len] = '\0'; 55 dig[i].st = sy[i].st = l[i].st = -1; 56 dig[i].ed = sy[i].ed = l[i].ed = i; 57 for (int j = 0, k = len; j <= k; j ++, k --){ 58 if (l[i].st == -1 && (is_l(s[j]) || is_l(s[k]))) 59 l[i].st = j; 60 if (dig[i].st == -1 && (is_dig(s[j]) || is_dig(s[k]))) 61 dig[i].st = j; 62 if (sy[i].st == -1 && (is_sy(s[j]) || is_sy(s[k]))) 63 sy[i].st = j; 64 65 if (~dig[i].st && ~sy[i].st && ~l[i].st) 66 break; 67 } 68 } 69 70 sort(l, l + n); 71 sort(dig, dig + n); 72 sort(sy, sy + n); 73 74 cout << get_ans(); 75 }
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:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 //#include<iostream> 5 #include<cmath> 6 #include<queue> 7 #include<map> 8 #include<string> 9 #include<vector> 10 using namespace std; 11 #define INF 2000000000 12 #define Clear(x, Num) memset(x, Num, sizeof(x)) 13 #define Dig(x) ((x>='0') && (x<='9')) 14 #define Neg(x) (x=='-') 15 #define G_c() getchar() 16 #define Maxn 1000010 17 typedef long long ll; 18 19 struct Data 20 { 21 int rank, id; 22 ll l, r; 23 }c[Maxn]; 24 int n; 25 ll l, r, a[Maxn], b[Maxn]; 26 27 inline int gcd(int x, int y) { if (!y) return x; return gcd(y, x%y); } 28 inline void read(int &x){ char ch; int N=1; while ((ch=G_c()) && (!Dig(ch)) && (!Neg(ch))); if (Neg(ch)) { N=-1; while ((ch=G_c()) && (!Dig(ch))); } x=ch-48; while ((ch=G_c()) && (Dig(ch))) x=x*10+ch-48; x*=N; } 29 //inline void Insert(int u, int v) { To[Cnt]=v; Next[Cnt]=Head[u]; Head[u]=Cnt++; } 30 inline bool cmp(const Data &a, const Data &b) { return a.rank<b.rank; } 31 int main() 32 { 33 scanf("%d%lld%lld", &n, &l, &r); 34 for (int i=1; i<=n; i++) scanf("%lld", &a[i]); 35 for (int i=1; i<=n; i++) scanf("%d", &c[i].rank), c[i].id=i; 36 sort(c+1, c+n+1, cmp); 37 c[1].l=l-a[c[1].id]; c[1].r=r-a[c[1].id]; 38 for (int i=2; i<=n; i++) 39 { 40 c[i].l=max(l-a[c[i].id], c[i-1].l+1); 41 c[i].r=min(r-a[c[i].id], c[i-1].r+1); 42 if (c[i].l>c[i].r) { puts("-1"); return 0; } 43 } 44 b[c[n].id]=c[n].l; 45 for (int i=n-1; i>=1; i--) 46 { 47 c[i].r=min(c[i].r, b[c[i+1].id]); 48 if (c[i].l>c[i].r) { puts("-1"); return 0; } 49 b[c[i].id]=c[i].l; 50 } 51 for (int i=1; i<=n; i++) printf("%lld ", b[i]+a[i]); puts(""); 52 }
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。
1 #define PRON "e" 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 #define inf 0x3f3f3f3f 8 #define LL "%lld" 9 #define st first 10 #define nd second 11 using namespace std; 12 typedef long long ll; 13 14 #define mp make_pair 15 16 const int maxn = 30 + 10; 17 const int ax[] = {-1, 0, 1, 0}; 18 const int ay[] = {0, 1, 0, -1}; 19 20 vector<int> g[maxn]; 21 pair<ll, ll> ans[maxn]; 22 int n; 23 24 inline void fail(){ 25 printf("NO"); 26 exit(0); 27 } 28 29 void dfs(int u, int pre, ll len, int dir, ll x, ll y){ 30 if (g[u].size() > 4) 31 fail(); 32 33 ans[u] = mp(x, y); 34 for (int i = 0, v, now_dir = 0; i < g[u].size(); i ++){ 35 v = g[u][i]; 36 if (v == pre) 37 continue; 38 39 if (pre != 0 && (dir + 2) % 4 == now_dir) 40 now_dir ++; 41 42 dfs(v, u, len / 2, now_dir, x + len * (ll)ax[now_dir], y + len * (ll)ay[now_dir]); 43 now_dir ++; 44 } 45 } 46 47 48 int main(){ 49 #ifndef ONLINE_JUDGE 50 freopen(PRON ".in", "r", stdin); 51 #endif 52 53 cin >> n; 54 for (int i = 0, u, v; i < n - 1; i ++){ 55 cin >> u >> v; 56 g[u].push_back(v); 57 g[v].push_back(u); 58 } 59 60 dfs(1, 0, 1ll << 35, 0, 0ll, 0ll); 61 62 cout << "YES\n"; 63 for (int i = 1; i <= n; i ++) 64 cout << ans[i].st << " " << ans[i].nd << endl; 65 }
Problems:
Analysis:
