CodeForces 401D Roman and Numbers

本文介绍了一种使用动态规划(DP)算法解决构造特定条件数字的问题,详细解释了算法思路、状态转移方程和注意事项,包括去重处理,最终通过代码实现了解决方案。

题意:

将n的各位数字重新排列(不允许有前导零)  求  可以构造几个%m等于0的数字


思路:

dp题  由于最多只有18个数字所以可以用二进制表示  dp[i][j]表示使用了i(二进制)表示的数字%m等于j的个数

状态转移  dp[i][ (k*10 + dig[j]) %y ]+=dp[ i ^ (1<<j) ][k]

注意去重  如果一个数字有x个  那么对于这个数字就产生了x!的重复


代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

__int64 x;
__int64 dp[1<<18][100],a[20];
int dig[20],num[10];
int y,len;

int main()
{
    int i,j,k;
    scanf("%I64d%d",&x,&y);
    while(x)
    {
        i=x%10;
        x/=10;
        dig[len++]=i;
        num[i]++;
    }
    sort(dig,dig+len);
    for(i=0;i<len;i++)
        if(dig[i])
            dp[1<<i][dig[i]%y]=1;
    for(a[0]=1,i=1;i<=len;i++) a[i]=a[i-1]*i;
    for(i=1;i<(1<<len);i++)
    {
        for(j=0;j<len;j++)
        {
            if(i&(1<<j))
            {
                for(k=0;k<y;k++)
                {
                    dp[i][(k*10+dig[j])%y]+=dp[i^(1<<j)][k];
                }
            }
        }
    }
    //for(i=1;i<(1<<len);i++)
    //    for(k=0;k<y;k++) printf("%d %d %I64d\n",i,k,dp[i][k]);
    for(i=0;i<10;i++) dp[(1<<len)-1][0]/=a[num[i]];
    printf("%I64d\n",dp[(1<<len)-1][0]);
    return 0;
}


### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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