第十八周 12.27-1.2

因为期末了本来不想写了 但是觉得挂篇空文也无所谓 还是写了

 

 

 

12.27

去西工大玩。

 

12.28-12.29

什么都没干。

 

12.30

上次挂的BC。

HDU 5602 Black Jack

一开始以为A用1表示。

WA了pretest。

看clar改完过了pre然后fst。

一直以为是dp错。找了很久很久找不出来。

后来看div2的clar才知道只有10用T表示。JQK还是JQK。

这样都能过pre。sample也没有。无语了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 double dp1[22][22], dp2[22][22];//x z
 7 double p[11] = { 0,
 8     1.0 / 13, 1.0 / 13, 1.0 / 13,
 9     1.0 / 13, 1.0 / 13, 1.0 / 13,
10     1.0 / 13, 1.0 / 13, 1.0 / 13, 4.0 / 13
11 };
12 
13 void pre()
14 {
15     for(int i = 2; i <= 21; i++)
16         for(int j = i; j <= 21; j++)
17             dp1[i][j] = 1;
18 
19     for(int i = 21; i >= 2; i--)
20         for(int j = i - 1; j >= 2; j--)
21             for(int k = 10; k >= 1; k--)
22                 if(j + k <= 21) dp1[i][j] += p[k] * dp1[i][j+k];
23 
24     for(int i = 21; i >= 2; i--)
25     {
26         for(int j = 21; j >= 2; j--)
27         {
28             double tmp = 0;//draw->lose
29             for(int k = 10; k >= 1; k--)
30             {
31                 if(i + k > 21) tmp += p[k];
32                 else tmp += p[k] * dp2[i+k][j];
33             }
34             dp2[i][j] = min(dp1[i][j], tmp);
35         }
36     }
37     return;
38 }
39 
40 int num(char c)
41 {
42     if(c == 'A') return 1;
43     if(c >= '0' && c <= '9') return c - '0';
44     return 10;
45 }
46 
47 int main(void)
48 {
49     pre();
50     int T;
51     scanf("%d", &T);
52     while(T--)
53     {
54         int s1, s2;
55         char s[11];
56         scanf("%s", s);
57         s1 = num(s[0]) + num(s[1]);
58         s2 = num(s[2]) + num(s[3]);
59         puts(dp2[s1][s2] < 0.5 ? "YES" : "NO");
60     }
61     return 0;
62 }
Aguin

 

12.31

CF 611 D New Year and Ancient Prophecy

n2搞下LCP就好了。QAQ

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long LL;
 6 const LL mod = 1e9 + 7;
 7 LL dp[5005][5005];
 8 char num[5005];
 9 int com[5005][5005];
10 
11 bool bigger(int pos1, int pos2, int len)
12 {
13     int x = com[pos1][pos2];
14     if( x >= len) return false;
15     return num[pos1+x] > num[pos2+x];
16 }
17 
18 int main(void)
19 {
20     int n;
21     scanf("%d%s", &n, num + 1);
22 
23     for(int i = n; i >= 1; i--)
24         for(int j = n; j >= 1; j--)
25             com[i][j] = num[i] == num[j] ? ( com[i+1][j+1] + 1 ) : 0 ;
26 
27     for(int i = 1; i <= n; i++)
28     {
29         for(int j = 1; j <= i; j++)
30         {
31             if(num[i-j+1] == '0') dp[i][j] = dp[i][j-1];
32             else if(i == j) dp[i][j] = ( dp[i][j-1] + 1LL ) % mod;
33             else if( i >= 2 * j && bigger(i-j+1, i-2*j+1, j) ) dp[i][j] = ( dp[i][j-1] + dp[i-j][j] ) % mod;
34             else dp[i][j] = ( dp[i][j-1] + dp[i-j][min(i-j,j-1)]) % mod;
35         }
36     }
37     printf("%I64d\n", dp[n][n]);
38     return 0;
39 }
Aguin

 

1.1

CF 611 E New Year and Three Musketeers

多重集搞搞。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <set>
 5 using namespace std;
 6 multiset<int> M;
 7 multiset<int>::iterator it;
 8 int m[3];
 9 
10 int main(void)
11 {
12     int n;
13     scanf("%d", &n);
14     for(int i = 0; i < 3; i++) scanf("%d", m + i);
15     sort(m, m + 3);
16     int a = m[0], b = m[1], c = m[2];
17 
18     for(int i = 0; i < n; i++)
19     {
20         int t;
21         scanf("%d", &t);
22         if(t > a + b + c) {puts("-1"); return 0;}
23         M.insert(t);
24     }
25 
26     int ans = 0;
27 
28     while(!M.empty())
29     {
30         it = M.lower_bound(b + c + 1);
31         if(it == M.end()) break;
32         M.erase(it); ans++;
33     }
34 
35     while(!M.empty())
36     {
37         it = M.lower_bound(a + c + 1);
38         if(it == M.end()) break;
39         M.erase(it); ans++;
40         if(M.empty()) break;
41         it = M.lower_bound(a + 1);
42         if(it != M.begin()) M.erase(--it);
43     }
44 
45     while(!M.empty())
46     {
47         it = M.lower_bound(max(c, a + b) + 1);
48         if(it == M.end()) break;
49         M.erase(it); ans++;
50         if(M.empty()) break;
51         it = M.lower_bound(b + 1);
52         if(it != M.begin()) M.erase(--it);
53     }
54 
55     int x = 0, y = 0, k = 0;
56     for(it = M.begin(); it != M.end(); it++)
57     {
58         if(*it <= a + b) x++;
59         if(*it <= c) y++;
60     }
61 
62     int s = max( (max(x, y) + 1) /2, max(x-y, y-x) );
63     while(!M.empty() && y)
64     {
65         it = M.lower_bound(c + 1);
66         if(it != M.begin())
67         {
68             --it;
69             if(*it <= a + b) x--;
70             if(*it <= c) y--;
71             M.erase(it);
72         }
73         it = M.lower_bound(b + 1);
74         if(it != M.begin())
75         {
76             --it;
77             if(*it <= a + b) x--;
78             if(*it <= c) y--;
79             M.erase(it);
80         }
81         it = M.lower_bound(a + 1);
82         if(it != M.begin())
83         {
84             --it;
85             if(*it <= a + b) x--;
86             if(*it <= c) y--;
87             M.erase(it);
88         }
89         s = min( s, ++k + max( (max(x, y) + 1)/2, max(x-y, y-x) ) );
90     }
91 
92     printf("%d\n", ans + s);
93 
94     return 0;
95 }
Aguin

 

1.2

矩阵快速幂！矩阵快速幂！矩阵快速幂！

一直懒得自己码板。每次都是用到去搜。受不了了！

 1 struct Matrix
 2 {
 3     LL m[maxn][maxn];
 4     Matrix(){memset(m, 0, sizeof(m));}
 5     void E(){for(int i = 0; i < maxn; i++) m[i][i] = 1;}
 6 };
 7 
 8 Matrix M_mul(Matrix a, Matrix b)
 9 {
10     Matrix ret;
11     for(int i = 0; i < maxn; i++)
12         for(int j = 0; j < maxn; j++)
13             for(int k = 0; k < maxn; k++)
14                 ret.m[i][j] = ( ret.m[i][j] + (a.m[i][k] * b.m[k][j]) % mod ) % mod;
15     return ret;
16 }
17 
18 Matrix M_qpow(Matrix P, LL n)
19 {
20     Matrix ret;
21     ret.E();
22     while(n)
23     {
24         if(n & 1LL) ret = M_mul(ret, P);
25         n >>= 1LL;
26         P = M_mul(P, P);
27     }
28     return ret;
29 }
Aguin

 

HDU 5607 graph

直接搞！

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 typedef long long LL;
  7 const int maxn = 55;
  8 const LL mod = 1e9 + 7;
  9 int G[maxn][maxn], deg[maxn];
 10 LL ans[22][maxn];
 11 
 12 struct query
 13 {
 14     int id, u;
 15     LL K;
 16 }q[22];
 17 
 18 bool cmp(query a, query b)
 19 {
 20     return a.K < b.K;
 21 }
 22 
 23 LL qpow(LL a, LL b)
 24 {
 25     LL ret = 1LL, tmp = a;
 26     while(b)
 27     {
 28        if(b & 1LL) ret = ret * tmp % mod;
 29        tmp = tmp * tmp % mod;
 30        b >>= 1LL;
 31     }
 32     return ret;
 33 }
 34 
 35 LL inv(LL a)
 36 {
 37     LL k = 1e9 + 5;
 38     return qpow(a, k);
 39 }
 40 
 41 struct Matrix
 42 {
 43     LL m[maxn][maxn];
 44     Matrix(){memset(m, 0, sizeof(m));}
 45     void E(){for(int i = 0; i < maxn; i++) m[i][i] = 1;}
 46 };
 47 
 48 Matrix M_mul(Matrix a, Matrix b)
 49 {
 50     Matrix ret;
 51     for(int i = 0; i < maxn; i++)
 52         for(int j = 0; j < maxn; j++)
 53             for(int k = 0; k < maxn; k++)
 54                 ret.m[i][j] = ( ret.m[i][j] + (a.m[i][k] * b.m[k][j]) % mod ) % mod;
 55     return ret;
 56 }
 57 
 58 Matrix M_qpow(Matrix P, LL n)
 59 {
 60     Matrix ret;
 61     ret.E();
 62     while(n)
 63     {
 64         if(n & 1LL) ret = M_mul(ret, P);
 65         n >>= 1LL;
 66         P = M_mul(P, P);
 67     }
 68     return ret;
 69 }
 70 
 71 int main(void)
 72 {
 73     int N, M;
 74     scanf("%d%d", &N, &M);
 75     for(int i = 1; i <= M; i++)
 76     {
 77         int X, Y;
 78         scanf("%d%d", &X, &Y);
 79         deg[X]++;
 80         G[X][Y] = 1;
 81     }
 82     int Q;
 83     scanf("%d", &Q);
 84     for(int i = 0; i < Q; i++)
 85     {
 86         scanf("%d%I64d", &q[i].u, &q[i].K);
 87         q[i].id = i;
 88     }
 89     sort(q, q + Q, cmp);
 90     Matrix A, B = M_qpow(A, 0LL);
 91     for(int i = 1; i <= N; i++)
 92     {
 93         LL R = inv(deg[i]);
 94         for(int j = 1; j <= N; j++)
 95             if(G[i][j]) A.m[j][i] = R;
 96     }
 97     for(int i = 0; i < Q; i++)
 98     {
 99         LL nn;
100         if(!i) nn = q[i].K;
101         else nn = q[i].K - q[i-1].K;
102         B = M_mul(B, M_qpow(A, nn));
103         for(int j = 1; j <= N; j++) ans[q[i].id][j] = B.m[j][q[i].u];
104     }
105     for(int i = 0; i < Q; i++)
106     {
107         for(int j = 1; j <= N; j++) printf("%d ", ans[i][j]);
108         puts("");
109     }
110     return 0;
111 }
Aguin

 

posted @ 2015-12-30 20:11  Aguin  阅读(184)  评论(0)  编辑  收藏  举报