codeforces 1073E

本文介绍了一种使用数位动态规划解决特定问题的方法。通过状态压缩记录已出现的数字,并定义函数f和g来分别计算可能的状态数及这些状态下的数字总和。此方法能够有效地解决数字范围内的相关计数问题。

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题解:

考虑数位DP,状压出现过的数字集合S,f ( l , x , S , pz , lim )表示到第 l 位,数字为x, 数字集合为S ,是否为前导0,是否贴上界

然后同时定义g为该状态下的数字和,利用 10^(l-1) * f(l , x, S, pz, lim)计算该位的贡献,然后加上所有后继的g就行了

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const ll mod = 998244353;
 5 ll l,r;
 6 int k;
 7 int num[22],cnt;
 8 ll f[22][10][1050][2][2],g[22][10][1050][2][2];
 9 bool vis[22][10][1050][2][2];
10 ll fastpow(ll a,ll p)
11 {
12     ll ans=1;
13     while(p)
14     {
15         if(p&1)ans=ans*a%mod;
16         a=a*a%mod;p>>=1;
17     }
18     return ans;
19 }
20 void dfs(int l,int x,int S,bool pz,bool lim)
21 {
22     if(vis[l][x][S][pz][lim])return;
23     vis[l][x][S][pz][lim]=1;
24     int t=0;
25     for(int i=0;i<10;++i)if(S&(1<<i))t++;
26     if(t>k)return;
27     if(l==1)
28     {
29         f[l][x][S][pz][lim]=1,g[l][x][S][pz][lim]=x;
30         return;
31     }
32     int up=(lim)?num[l-1]:9;
33     for(int i=0;i<=up;++i)
34     {
35         dfs(l-1,i,(pz&(!i))?0:(S|(1<<i)),pz&(!i),lim&(i==num[l-1]));
36         f[l][x][S][pz][lim]=(f[l][x][S][pz][lim]+f[l-1][i][(pz&(!i))?0:(S|(1<<i))][pz&(!i)][lim&(i==num[l-1])])%mod;
37         g[l][x][S][pz][lim]=(g[l][x][S][pz][lim]+g[l-1][i][(pz&(!i))?0:(S|(1<<i))][pz&(!i)][lim&(i==num[l-1])])%mod;
38     }
39     g[l][x][S][pz][lim]=(g[l][x][S][pz][lim]+f[l][x][S][pz][lim]*x%mod*fastpow(10,l-1)%mod)%mod;
40 }
41 ll solve(ll n)
42 {
43     if(!n)return 0; 
44     memset(num,0,sizeof(num));
45     memset(f,0,sizeof(f));
46     memset(g,0,sizeof(g));
47     memset(vis,0,sizeof(vis));
48     cnt=0;
49     ll x=n;
50     while(x)
51     {
52         num[++cnt]=x%10;
53         x/=10;
54     }
55     ll ans=0;
56     for(int i=0;i<=num[cnt];++i)
57     {
58         dfs(cnt,i,(i==0)?0:(1<<i),(i==0),(i==num[cnt]));
59         ans=(ans+g[cnt][i][(i==0)?0:(1<<i)][(i==0)][(i==num[cnt])])%mod;
60     }
61     return ans;
62 }
63 int main()
64 {
65     scanf("%I64d%I64d%d",&l,&r,&k);
66     printf("%I64d\n",(solve(r)-solve(l-1)+mod)%mod);
67     return 0;
68 }
View Code

 

转载于:https://www.cnblogs.com/uuzlove/p/10612406.html

### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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