正睿20NOIp前冲刺day1

估分：100+10+40+30=180

实际：60+0+40+20=120

T1:

　　签到题，有手就行，但我用了cin，超时了

　　垃圾cin，再也不用了

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<map>
 4 #include<iostream>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 typedef unsigned long long ULL;
 9 
10 const int N = 10000010;
11 
12 char str1[N], str2[N];
13 map<char, int> st;
14 
15 int main()
16 {
17     scanf("%s%s", str1, str2);
18     int n = strlen(str1), m = strlen(str2);
19     ULL ans = (ULL)(n + 1) * (m + 1);
20     for (int i = 0; i < n; i++) st[str1[i]]++;
21     for (int i = 0; i < m; i++) ans -= st[str2[i]];
22     printf("%llu", ans);
23 }

View Code

T2:

　　暴力没写好，10分没了

　　当a全部相等的时候，用倍增DP，f(i,j)表示前2ⁱ种糖果，占了j个位置的方案数。转移方程如下

　　初始化，g(0,0...ai)=1。

　　如果a不相等，把≥m的a全部变成m，对于所有种类的a分别做一遍上面的DP，然后合并起来。

　　因为n很大，所以用多项式快速幂来做

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<vector>
  4 #include<iostream>
  5 #include<algorithm>
  6 using namespace std;
  7 
  8 typedef long long LL;
  9 
 10 const int N = 105, M = 10000005;
 11 const int mod = 998244353;
 12 
 13 int vis[M], m;
 14 LL n;
 15 LL cnt[N], f[N], g[N], fac[N], inv[N];
 16 
 17 LL qmul(LL a, int k) 
 18 {
 19     LL res = 1;
 20     while(k)
 21     {
 22         if (k & 1) res = res * a % mod;
 23         a = a * a % mod;
 24         k >>= 1;
 25     }
 26     return res;
 27 }
 28 
 29 void init()
 30 {
 31     fac[0] = 1;
 32     for (int i = 1; i <= 100; i++) fac[i] = fac[i - 1] * i % mod;
 33     inv[100] = qmul(fac[100], mod - 2);
 34     for (int i = 100; i; i--) inv[i - 1] = inv[i] * i % mod;
 35 }
 36 
 37 void mul(LL* a, const LL* b) 
 38 {
 39     LL c[N];
 40     memset(c, 0, sizeof(c));
 41     for (int i = 0; i <= m; i++)
 42     {
 43         for (int j = 0; i + j <= m; j++)
 44         {
 45             c[i + j] = (c[i + j] + a[i] * b[j]) % mod;
 46         }
 47     }    
 48     for (int i = 0; i <= m; i++) a[i] = c[i];
 49 }
 50 
 51 void calc(int p, LL k) 
 52 {
 53     LL t[N];
 54     memset(t, 0, sizeof(t));
 55     memset(g, 0, sizeof(g));
 56 
 57     g[0] = t[0] = 1;
 58     for (int i = 0; i <= p; i++) t[i] = inv[i];
 59     while (k)
 60     {
 61         if (k & 1) mul(g, t);
 62         mul(t, t);
 63         k >>= 1;
 64     }
 65 }
 66 
 67 int main() 
 68 {
 69     init();
 70     scanf("%lld%d", &n, &m);
 71 
 72     int  s, p;
 73     int a1, A, B, P;
 74     scanf("%d%d%d%d", &a1, &A, &B, &P);
 75 
 76     vis[a1] = 1;
 77     for (int i = 2; ; i++)
 78     {
 79         a1 = ((LL)a1 * A + B) % P + 1;
 80         if (vis[a1])
 81         {
 82             s = vis[a1] - 1;
 83             p = i - vis[a1];
 84             break;
 85         }
 86         vis[a1] = i;
 87     }
 88 
 89     if (!p) s = n;
 90     LL t = (n - s) / p, q = (n - s) % p;
 91     for (int i = 1; i <= P; i++)
 92     {
 93         if (vis[i])
 94         {
 95             if (vis[i] <= s) ++cnt[min(m, i)];
 96             else cnt[min(m, i)] += t + (vis[i] - s <= q);
 97         }
 98     }
 99 
100     f[0] = 1;
101     for (int i = 1; i <= m; i++) 
102     {
103         calc(i, cnt[i]);
104         mul(f, g);
105     }
106     printf("%lld\n", f[m] * fac[m] % mod);
107 }

View Code

T3：

　　|T|≤2的没写完

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 #include<iostream>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 const int N = 100010;
 9 
10 int n, m;
11 int nex[N][26], f[26][52];
12 char str1[N], str2[N];
13 
14 int main()
15 {
16     scanf("%s %s", str1 + 1, str2 + 1);
17     n = strlen(str1 + 1), m = strlen(str2 + 1);
18 
19     for (int i = 0; i < 26; i++) nex[n + 1][i] = n + 1;
20     for (int i = n; ~i; i--) 
21     {
22         for (int j = 0; j < 26; j++) 
23         {
24             nex[i][j] = nex[i + 1][j];
25             if (i < n && str1[i + 1] == j + 'a') nex[i][j] = i + 1;
26         }
27     }
28 
29     int q;
30     scanf("%d", &q);
31     while(q--)
32     {
33         int l, r;
34         scanf("%d %d", &l, &r);
35         memset(f, 0x3f, sizeof(f));
36 
37         f[0][m] = l - 1;
38         for (int j = 1; j <= m; j++) 
39         {
40             for (int k = 0; k <= 2 * m; k++)
41             {
42                 if (f[j - 1][k] <= n)
43                 {
44                     if (k) f[j][k - 1] = min(f[j][k - 1], f[j - 1][k]);
45                     f[j][k] = min(f[j][k], f[j - 1][k] + 1);
46                     f[j][k + 1] = min(f[j][k + 1], nex[f[j - 1][k]][str2[j] - 'a']);
47                 }
48             }
49         }
50 
51         for (int j = 2 * m; ~j; j--)
52         {
53             if (f[m][j] <= r)
54             {
55                 printf("%d\n", r - l + 1 - (j - m));
56                 break;
57             }
58         }
59     }
60 }

View Code

T4:　

　　有10分的暴力写炸了

　　不会

总结：

·　　动态规划还是弱项，要多锻炼锻炼。以后再也不用cin了，慢的要死。

posted on 2020-10-12 17:28 ArrogHie 阅读(125) 评论(0) 编辑收藏举报