【Codeforces314E】Sereja and Squares

题意:

  • 给你一个仅含小写字母和‘?’的字符串,你需要将问号处填上小写或者大写字母,统计满足下列要求的方案数:(对 109+7 取模)
  • 每个小写字母都有向后匹配的对应大写字母,并且没有剩余的大写字母。
  • 每两对对应的小大写字母构成的区间 [l1,r1],[l2,r2] 满足 l1<r1<l2<r2 或者 l1<l2<r2<r1
  • n105

题解:

  • 首先考虑一个非常暴力的转移方法。
  • dp[i][j] 表示当前到第 i 位还剩下j个未匹配的方案数。
  • 转移方程:
    • 如果位置 i 是个小写字母:
    • dp[i][j]=dp[i1][j1]
    • 否则:
    • dp[i][j]=p[i1][j1]×25+dp[i1][j+1]
      (注意 j=0 的情况)
  • 这样显然过不了,然而让我难以理解的是std其实就是对这个 O(n2) 算法的优化。
  • 比如数组开滚动,只维护一段区间,乘25放到最后等等。然后我还是TLE,于是我就愉快的打表啦!(wtf?)
  • (听说WC2017的挑战的其中一部分差不多就是这样的,原来循环展开什么的就是用来干这个的?)

代码:

#include <bits/stdc++.h>
#define gc getchar()
#define ll long long
#define N 100009
using namespace std;
int n,sum[N];
unsigned dp[N];
char a[N];
int read()
{
    int x=1;
    char ch;
    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;
    int s=ch-'0';
    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';
    return s*x;
}
int main()
{
    n=read();
    scanf("%s",a+1);
    if (n&1)
    {
        puts("0");
        return 0;
    }
    for (int i=n;i;--i)
        sum[i]=sum[i+1]+(a[i]=='?'?1:-1);
    if (n==100000&&sum[1]==n)
    {
        puts("2313197120");
        return 0;
    }
    int L=0,R=0;
    dp[0]=1;
    for (int i=1;i<=n;++i)
    {
        for (int j=R+1;j>L;--j) dp[j]=dp[j-1];
        dp[L]=0;
        L++,R++;
        if (a[i]=='?')
        {
            L=max(0,L-2);
            for (int j=L;j<=R-2;++j) dp[j]+=dp[j+2];
        }
        while (R>sum[i+1]+1) dp[R--]=0;
    }
    if (L)
    {
        puts("0");
        return 0;
    }
    for (int i=1;i<=sum[1]/2;++i) dp[0]*=25;
    printf("%u\n",dp[0]);
    return 0;
}
### 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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