MemSQL Start[c]UP 2.0 - Round 1

搞了好久才把大部分题目题解看完了，真是太弱了。

A题简单暴力题一个一个匹配，对应位置字母要么相同，要么是'.'.

B题给定一个矩阵，左下角(0,0)，右上角(n, m)，取4个不同的点连成一段折线，要有最长的折线长度。

排除n == 0 和m == 0 ，剩下的情况中总共由4中情况：

枚举一下就可以了

1. (0,0)->(n,m)->(0, m)->(n, 0)

2. (0,0)->(n,m)->(n, 0)->(0, m)

3.(n,m-1)->(0,0)->(n,m)->(0,1)

4.(0,m-1)->(n,0)->(0,m)->(n,1)

代码：

  1 //Template updates date: 20140718
  2 #include <iostream>
  3 #include <sstream>
  4 #include <cstdio>
  5 #include <climits>
  6 #include <ctime>
  7 #include <cctype>
  8 #include <cstring>
  9 #include <cstdlib>
 10 #include <string>
 11 #include <stack>
 12 #include <set>
 13 #include <map>
 14 #include <cmath>
 15 #include <vector>
 16 #include <queue>
 17 #include <algorithm>
 18 #define  esp 1e-6
 19 #define  inf 0x3f3f3f3f
 20 #define  pi acos(-1.0)
 21 #define  pb push_back
 22 #define  lson l, m, rt<<1
 23 #define  rson m+1, r, rt<<1|1
 24 #define  lowbit(x) (x&(-x))
 25 #define  mp(a, b) make_pair((a), (b))
 26 #define  bit(k) (1<<(k))
 27 #define  iin  freopen("pow.in", "r", stdin);
 28 #define  oout freopen("pow.out", "w", stdout);
 29 #define  in  freopen("solve_in.txt", "r", stdin);
 30 #define  out freopen("solve_out.txt", "w", stdout);
 31 #define  bug puts("********))))))");
 32 #define  Inout iin oout
 33 #define  inout in out
 34 
 35 #define  SET(a, v) memset(a, (v), sizeof(a))
 36 #define  SORT(a)   sort((a).begin(), (a).end())
 37 #define  REV(a)    reverse((a).begin(), (a).end())
 38 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
 39 #define  REP(i, n) for(int i = 0; i < (n); i++)
 40 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
 41 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
 42 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
 43 #define  pf(x) ((x)*(x))
 44 #define  mod(n) ((n))
 45 #define  Log(a, b) (log((double)b)/log((double)a))
 46 #define Srand() srand((int)time(0))
 47 #define random(number) (rand()%number)
 48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
 49 
 50 using namespace std;
 51 typedef long long  LL;
 52 typedef unsigned long long ULL;
 53 typedef vector<int> VI;
 54 typedef pair<int,int> PII;
 55 typedef vector<PII> VII;
 56 typedef vector<PII, int> VIII;
 57 typedef VI:: iterator IT;
 58 typedef map<string, int> Mps;
 59 typedef map<int, int> Mpi;
 60 typedef map<int, PII> Mpii;
 61 typedef map<PII, int> Mpiii;
 62 double ans[4];
 63 
 64 int main() {
 65 
 66     int n, m;
 67     cin>>n>>m;
 68     if(n == 0) {
 69         printf("%d %d\n", 0, 1);
 70         printf("%d %d\n", 0, m);
 71         printf("%d %d\n", 0, 0);
 72         printf("%d %d\n", 0, m-1);
 73     } else if(m == 0) {
 74         printf("%d %d\n", 1, 0);
 75         printf("%d %d\n", n, 0);
 76         printf("%d %d\n", 0, 0);
 77         printf("%d %d\n", n-1, 0);
 78     } else {
 79         ans[0] = m + 2*sqrt(pf(n)+pf(m));
 80         ans[1] = n + 2*sqrt(pf(n)+pf(m));
 81         ans[2] = sqrt(pf(n)+pf(m))+sqrt(pf(n)+pf(m-1))*2;
 82         ans[3] = sqrt(pf(n)+pf(m))+sqrt(pf(n-1)+pf(m))*2;
 83         int tmp = 0;
 84         REP(i, 4)
 85         if(ans[tmp] < ans[i])
 86             tmp = i;
 87         if(tmp == 0) {
 88             printf("%d %d\n", 0, 0);
 89             printf("%d %d\n", n, m);
 90             printf("%d %d\n", n, 0);
 91             printf("%d %d\n", 0, m);
 92         } else if(tmp == 1) {
 93             printf("%d %d\n", 0, 0);
 94             printf("%d %d\n", n, m);
 95             printf("%d %d\n", 0, m);
 96             printf("%d %d\n", n, 0);
 97         } else if(tmp == 3) {
 98             printf("%d %d\n", 1, m);
 99             printf("%d %d\n", n, 0);
100             printf("%d %d\n", 0, m);
101             printf("%d %d\n", n-1, 0);
102         } else {
103             printf("%d %d\n", n, m-1);
104             printf("%d %d\n", 0, 0);
105             printf("%d %d\n", n, m);
106             printf("%d %d\n", 0, 1);
107         }
108     }
109     return 0;
110 }

View Code

C题:有n种卡片，每种m张,魔术师每次从中选取n张然后给观众选取，自己再从中选出一张，问最后选得的和观众相同的概率是多少？

分析：由于最后选得的相同的卡片肯定是n种中一种，而且是哪种不重要，这样只需考虑，该种卡片在选出的n张牌占多少？

选出卡片中该种有i张，然后最后魔术师和观众选中该种卡片，这样概率就是：C(m,i)*C((n-1)m, n-i)*(i/n)^2/C(nm,n) 1<=i<=min(n, m)

由于对应n种卡片都是上面这样计算的，所以上面结果*n就是最后结果了。

最后计算时候可以用log这样乘除法转化为+，-,对于概率计算很方便，对结果取一下e^ans放就可以了。

 

 1 //Template updates date: 20140718
 2 #include <iostream>
 3 #include <sstream>
 4 #include <cstdio>
 5 #include <climits>
 6 #include <ctime>
 7 #include <cctype>
 8 #include <cstring>
 9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define  esp 1e-6
19 #define  inf 0x3f3f3f3f
20 #define  pi acos(-1.0)
21 #define  pb push_back
22 #define  lson l, m, rt<<1
23 #define  rson m+1, r, rt<<1|1
24 #define  lowbit(x) (x&(-x))
25 #define  mp(a, b) make_pair((a), (b))
26 #define  bit(k) (1<<(k))
27 #define  iin  freopen("pow.in", "r", stdin);
28 #define  oout freopen("pow.out", "w", stdout);
29 #define  in  freopen("solve_in.txt", "r", stdin);
30 #define  out freopen("solve_out.txt", "w", stdout);
31 #define  bug puts("********))))))");
32 #define  Inout iin oout
33 #define  inout in out
34 
35 #define  SET(a, v) memset(a, (v), sizeof(a))
36 #define  SORT(a)   sort((a).begin(), (a).end())
37 #define  REV(a)    reverse((a).begin(), (a).end())
38 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define  REP(i, n) for(int i = 0; i < (n); i++)
40 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define  pf(x) ((x)*(x))
44 #define  mod(n) ((n))
45 #define  Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49 
50 using namespace std;
51 typedef long long  LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62 double nCr(int n, int m) {
63     double res  = 0;
64     Rep(i, 1, m+1)
65     res += -log(i)+log(n-i+1);
66     return res;
67 }
68 void solve(int n, int m) {
69     int M = min(n, m);
70     double ans = 0;
71     Rep(i, 1, M+1) {
72         double tmp = 2*(log(i)-log(n))+log(n)+nCr(m, i)+nCr((n-1)*m, n-i)-nCr(n*m, n);
73         ans += exp(tmp);
74     }
75     printf("%.9f\n", ans);
76 }
77 int main() {
78         int n, m;
79     cin>>n>>m;
80     solve(n, m);
81     return 0;
82 }

View Code

D题：有k件衣服，每件必须先后并且连续经过3种操作，洗涤，烘干，折叠，分别由3种机器来完成，其中机器数目为n1,n2,n3,一个机器只能同时对一件衣服进行对应操作。其中衣服完成每种操作的时间为t1,t2,t3.求最后完成所有衣服3种操作的最短时间。

分析：由于每种操作最多可能处理ni件衣服所以第ni+1件衣服和第1件衣服的开始处理时间必须相差ti，根据这个可以推理最终完成时间。

代码：

//Template updates date: 20140718
#include <bits/stdc++.h>
#define  esp 1e-6
#define  inf 0x3f3f3f3f
#define  pi acos(-1.0)
#define  pb push_back
#define  lson l, m, rt<<1
#define  rson m+1, r, rt<<1|1
#define  lowbit(x) (x&(-x))
#define  mp(a, b) make_pair((a), (b))
#define  bit(k) (1<<(k))
#define  iin  freopen("pow.in", "r", stdin);
#define  oout freopen("pow.out", "w", stdout);
#define  in  freopen("solve_in.txt", "r", stdin);
#define  out freopen("solve_out.txt", "w", stdout);
#define  bug puts("********))))))");
#define  Inout iin oout
#define  inout in out

#define  SET(a, v) memset(a, (v), sizeof(a))
#define  SORT(a)   sort((a).begin(), (a).end())
#define  REV(a)    reverse((a).begin(), (a).end())
#define  READ(a, n) {REP(i, n) cin>>(a)[i];}
#define  REP(i, n) for(int i = 0; i < (n); i++)
#define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
#define  Rep(i, base, n) for(int i = (base); i < (n); i++)
#define  REPS(s, i) for(int i = 0; (s)[i]; i++)
#define  pf(x) ((x)*(x))
#define  mod(n) ((n))
#define  Log(a, b) (log((double)b)/log((double)a))
#define Srand() srand((int)time(0))
#define random(number) (rand()%number)
#define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)

using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<PII, int> VIII;
typedef VI:: iterator IT;
typedef map<string, int> Mps;
typedef map<int, int> Mpi;
typedef map<int, PII> Mpii;
typedef map<PII, int> Mpiii;
int n, n1, n2, n3, t1, t2, t3;
const int maxn = 10000 + 100;

int w[maxn];
int main() {
    
    scanf("%d%d%d%d%d%d%d", &n, &n1, &n2, &n3, &t1, &t2, &t3);
    Rep(i, 1, n+1) {
        int c = 0;
        if(i > n1) c = max(c, w[i-n1]+t1);
        if(i > n2) c = max(c, w[i-n2]+t2);
        if(i > n3) c = max(c, w[i-n3]+t3);
        w[i] = c;
        if(i == n)
            cout<<c+t1+t2+t3<<endl;
    }
    return 0;
}

View Code

 

 

E题：给定3个串s1,s2,s3,对每一个l(1<=l<=min(|s1|,|s2|,|s3|))求三元组(i,j,k)的个数，三元组是指满足

##：s1[i,i+l-1], s2[j,j+l-1],s3[k,k+l-1]是完全相同字符串的条件。

分析：表示知道用后缀数组处理，然后就不知道怎么做了，看了题解还是有点不明白，先留个坑，

UPD:终于吧这道题弄懂了。

分析：题目是要统计对于长度为l的字串，三元组的(i, j, k)的个数，其中满足上面##的条件。很巧妙的用到了并查集的，我们可以利用后缀数组求出h数组。

将LCP值相同的对应的h下标，也就是h值相同的下标集中存储，然后按照h从大到小,依次将h数组中位于相邻位置，且LCP>=当前h值的下标用并查集合并在一起。

对于并查集中根节点rt, cnt[0][rt], cnt[1][rt], cnt[2][rt]表示LCP大于当前的h值中，所有的子串中出自第1,2,3个的有多少个，每次合并之前减去之前的值，利用合并之后的cnt数组进行更新。

代码：

  1 #include <bits/stdc++.h>
  2 #define in freopen("solve_in.txt", "r", stdin);
  3 #define bug(x) printf("Line #%d: >>>>><<<\n", (x));
  4 typedef long long LL;
  5 
  6 using namespace std;
  7 const LL M = 1000000007;
  8 
  9 const int maxn = 3*(int)1e5+1000;
 10 int s[maxn], sa[maxn], t[maxn], t2[maxn], c[maxn], h[maxn], rk[maxn];
 11 int n;
 12 char str[3][maxn];
 13 LL cnt[3][maxn];
 14 int fa[maxn];
 15 int slen[3];
 16 int res[maxn];
 17 
 18 vector<int> g[maxn];
 19 
 20 void buildSa(int m)
 21 {
 22     int *x = t, *y = t2;
 23     for(int i = 0; i < m; i++) c[i] = 0;
 24     for(int i = 0; i < n; i++) c[x[i] = s[i]]++;
 25     for(int i = 1; i < m; i++) c[i] += c[i-1];
 26     for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
 27 
 28     for(int k = 1; k <= n; k <<= 1)
 29     {
 30         int p = 0;
 31         for(int i = n-k; i < n; i++) y[p++] = i;
 32 
 33         for(int i = 0; i < m; i++) c[i] = 0;
 34         for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;
 35         for(int i = 0; i < n; i++) c[x[y[i]]]++;
 36         for(int i = 1; i < m; i++) c[i] += c[i-1];
 37 
 38         for(int  i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
 39         swap(x, y);
 40         p = 1;
 41         x[sa[0]] = 0;
 42         for(int i = 1; i < n; i++)
 43             x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
 44         if(p >= n) break;
 45         m = p;
 46     }
 47 }
 48 void getHeight()
 49 {
 50     int j, k = 0;
 51     h[0] = 0;
 52     for(int i = 0; i < n; i++) rk[sa[i]] = i;
 53     for(int i = 0; i < n; i++)
 54     {
 55 
 56         if(k) k--;
 57         if(rk[i] == 0)
 58             continue;
 59         j = sa[rk[i]-1];
 60         while(s[i+k] == s[j+k]) k++;
 61         h[rk[i]] = k;
 62     }
 63 }
 64 int findset(int x)
 65 {
 66     return x == fa[x] ? x : fa[x] = findset(fa[x]);
 67 }
 68 void Union(int x, int y)
 69 {
 70     int fx = findset(x);
 71     int fy = findset(y);
 72     if(fx == fy)
 73         return;
 74     for(int i = 0; i < 3; i++)
 75         cnt[i][fx] += cnt[i][fy];
 76     fa[fy] = fx;
 77 }
 78 
 79 void init()
 80 {
 81     for(int i = 0; i < maxn; i++)
 82     {
 83         fa[i] = i;
 84     }
 85     int base = 0;
 86     for(int k = 0; k < 3; k++)
 87     {
 88         for(int j = 0; j < slen[k]; j++)
 89             cnt[k][j+base+k] = 1;
 90         base += slen[k];
 91     }
 92 }
 93 
 94 void solve()
 95 {
 96     LL ans = 0;
 97     for(int i = n-1; i > 0; i--)
 98     {
 99         int sz = g[i].size();
100         for(int j = 0; j < sz; j++)
101         {
102             int v = g[i][j];
103             int x = findset(sa[v-1]);
104             int y = findset(sa[v]);
105             ans -= cnt[0][x]*cnt[1][x]%M*cnt[2][x]%M;
106             if(ans < 0)
107                 ans += M;
108             ans -= cnt[0][y]*cnt[1][y]%M*cnt[2][y]%M;
109             if(ans < 0)
110                 ans += M;
111             Union(x, y);
112             x = findset(x);
113             ans = (ans+cnt[0][x]*cnt[1][x]%M*cnt[2][x]%M)%M;
114             
115         }
116 
117         res[i] = ans;
118     }
119 
120     int len = min(slen[0], min(slen[1], slen[2]));
121     for(int i = 1; i <= len; i++)
122         printf("%d%c", res[i], i == len ? '\n' : ' ');
123 }
124 int main()
125 {
126 
127     scanf("%s%s%s", str[0], str[1], str[2]);
128     for(int k = 0; k < 3; k++)
129     {
130         for(int i = 0; str[k][i]; i++)
131         {
132             s[n++] = str[k][i] - 'a' + 1;
133             slen[k]++;
134         }
135         s[n++] = 30+k;
136     }
137     s[n++] = 0;
138     init();
139     buildSa(60);
140     getHeight();
141     for(int i = 1; i < n; i++)
142         if(h[i])
143             g[h[i]].push_back(i);
144     solve();
145     return 0;
146 }

View Code

思路很类似的一道题：

 

F题：给定一个数组a[1...n]，为1....n的数的排列， n<=300000,是否存在两个数a，b，a+b/2在这数组中位于这两个数之间。实际就是判断是否三个数构成等差数列。

分析：这题要用线段树+hash来维护，首先容易知道，a,b,c构成等差那么 a，b之差与b,c之差是相等的，那么过程是这样的依次每次取出一个数a[i]，然后把a[i]和n-a[i]+1插入到两棵线段树中；如果不存在两个数与a[i]构成等差数列，例如数组中10个元素，对于2,5,8，5不能和2,8构成等差，2,8在数组中位置要么在5前面，要么在5后面，如果2,8在前面，那么将2,8以及10-2+1 = 9,10-8+1=3插入后，两棵线段树相应位置均会被增加相同元素，这样对应的线段树就应该都是序列2...8,那么比较两棵线段树相应区间hash值就可以了。同样对于2,8出现在5后面也可以判断。

一旦遇到hash值不同的情况说明存在等差数列。

代码：

 

  1 //Template updates date: 20140718
  2 #include <bits/stdc++.h>
  3 #define  esp 1e-6
  4 #define  inf 0x3f3f3f3f
  5 #define  pi acos(-1.0)
  6 #define  pb push_back
  7 #define  lson l, m, rt<<1
  8 #define  rson m+1, r, rt<<1|1
  9 #define  lowbit(x) (x&(-x))
 10 #define  mp(a, b) make_pair((a), (b))
 11 #define  bit(k) (1<<(k))
 12 #define  iin  freopen("pow.in", "r", stdin);
 13 #define  oout freopen("pow.out", "w", stdout);
 14 #define  in  freopen("solve_in.txt", "r", stdin);
 15 #define  out freopen("solve_out.txt", "w", stdout);
 16 #define  bug puts("********))))))");
 17 #define  Inout iin oout
 18 #define  inout in out
 19 
 20 #define  SET(a, v) memset(a, (v), sizeof(a))
 21 #define  SORT(a)   sort((a).begin(), (a).end())
 22 #define  REV(a)    reverse((a).begin(), (a).end())
 23 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
 24 #define  REP(i, n) for(int i = 0; i < (n); i++)
 25 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
 26 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
 27 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
 28 #define  pf(x) ((x)*(x))
 29 #define  mod(n) ((n))
 30 #define  Log(a, b) (log((double)b)/log((double)a))
 31 #define Srand() srand((int)time(0))
 32 #define random(number) (rand()%number)
 33 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
 34 
 35 using namespace std;
 36 typedef long long  LL;
 37 typedef unsigned long long ULL;
 38 typedef vector<int> VI;
 39 typedef pair<int,int> PII;
 40 typedef vector<PII> VII;
 41 typedef vector<PII, int> VIII;
 42 typedef VI:: iterator IT;
 43 typedef map<string, int> Mps;
 44 typedef map<int, int> Mpi;
 45 typedef map<int, PII> Mpii;
 46 typedef map<PII, int> Mpiii;
 47 
 48 const int maxn = 300000 + 100;
 49 const int L = 100003;
 50 
 51 LL b[maxn];
 52 
 53 int n;
 54 int a[maxn];
 55 LL ans;
 56 struct SegTree {
 57     LL h[maxn<<2];
 58     void build(int l, int r, int rt) {
 59         if(l == r) {
 60             h[rt] = 0;
 61             return;
 62         }
 63         int m = (l+r)>>1;
 64         build(lson);
 65         build(rson);
 66     }
 67     void PushUp(int l, int r, int rt) {
 68         int m = (l+r)>>1;
 69         int ll = r-m;
 70         h[rt] = h[rt<<1]*b[ll]+h[rt<<1|1];
 71     }
 72     void update(int l, int r, int rt, int pos) {
 73         if(pos == l && pos == r) {
 74             h[rt] = b[1];
 75             return;
 76         }
 77         int m = (l+r)>>1;
 78         if(pos <= m)
 79             update(lson, pos);
 80         else update(rson, pos);
 81         PushUp(l, r, rt);
 82     }
 83     void query(int L, int R, int l, int r, int rt) {
 84         if(L > R)
 85             return;
 86         if(L <= l && R >= r) {
 87             ans = ans*b[r-l+1]+h[rt];
 88             return;
 89         }
 90         int m = (l+r)>>1;
 91         if(L <= m)
 92             query(L, R, lson);
 93         if(R > m)
 94             query(L, R, rson);
 95     }
 96 } t1, t2;
 97 int main() {
 98 
 99     scanf("%d", &n);
100     b[0] = 1;
101     Rep(i, 1, n+1) {
102         scanf("%d", a+i);
103         b[i] = b[i-1]*L;
104     }
105     t1.build(1, n, 1);
106     t2.build(1, n, 1);
107     Rep(i, 1, n+1) {
108         int v = a[i];
109         int l = min(v-1, n-v);
110         ans = 0;
111         t1.query(v-l, v-1, 1, n, 1);
112         LL tmp = ans;
113         ans = 0;
114         t2.query(n-v-l+1, n-v, 1, n, 1);
115         if(tmp != ans) {
116             puts("YES");
117             return 0;
118         }
119         t1.update(1, n, 1, v);
120         t2.update(1, n, 1, n-v+1);
121     }
122     puts("NO");
123     return 0;
124 }

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