• HDU 6143 Killer Names【dp递推】【好题】【思维题】【阅读题】


    Killer Names

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 91    Accepted Submission(s): 44


    Problem Description
    > Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as the sole offspring of two Jedi Knights—Mallie and Kento Marek—who deserted the Jedi Order during the Clone Wars. Following the death of his mother, the young Marek's father was killed in battle by Darth Vader. Though only a child, Marek possessed an exceptionally strong connection to the Force that the Dark Lord of the Sith sought to exploit.
    >
    > When Marek died in 2 BBY, shortly after the formation of the Alliance, Vader endeavored to recreate his disciple by utilizing the cloning technologies of the planet Kamino. The accelerated cloning process—an enhanced version of the Kaminoan method which allowed for a rapid growth rate within its subjects—was initially imperfect and many clones were too unstable to take Marek's place as the Dark Lord's new apprentice. After months of failure, one particular clone impressed Vader enough for him to hope that this version might become the first success. But as with the others, he inherited Marek's power and skills at the cost of receiving his emotions as well, a side effect of memory flashes used in the training process.
    >
    > — Wookieepedia

    Darth Vader is finally able to stably clone the most powerful soilder in the galaxy: the Starkiller. It is the time of the final strike to destroy the Jedi remnants hidden in every corner of the galaxy.

    However, as the clone army is growing, giving them names becomes a trouble. A clone of Starkiller will be given a two-word name, a first name and a last name. Both the first name and the last name have exactly n characters, while each character is chosen from an alphabet of size m. It appears that there are m2n possible names to be used.

    Though the clone process succeeded, the moods of Starkiller clones seem not quite stable. Once an unsatisfactory name is given, a clone will become unstable and will try to fight against his own master. A name is safe if and only if no character appears in both the first name and the last name.

    Since no two clones can share a name, Darth Vader would like to know the maximum number of clones he is able to create.
     

    Input
    The First line of the input contains an integer T (T10), denoting the number of test cases. 

    Each test case contains two integers n and m (1n,m2000).
     

    Output
    For each test case, output one line containing the maximum number of clones Vader can create.

    Output the answer  mod 109+7
     

    Sample Input
    2 3 2 2 3
     

    Sample Output
    2 18
     

    Source
    有m个数,两个长度为n的空位,要使两个长为n的空位填满,使两个长为n的段不出现相同的数字,问有多少种填法。

    dp[i][j]表示长度为i选了j个颜色的可能性为多少,则直接dp

    最后的结果即为左边的结果乘右边与左边不同的结果

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define ms(x,y) memset(x,y,sizeof(x))
    using namespace std;
    
    typedef long long ll;
    
    const double pi = acos(-1.0);
    const int mod = 1e9 + 7;
    const int maxn = 2010;
    
    ll dp[maxn][maxn];
    
    ll quick_pow(ll a, ll n) {
        ll ans = 1;
        while (n) {
            if (n & 1)    ans = ans*a%mod;
            a = a*a%mod;
            n >>= 1;
        }
        return ans;
    }
    
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int n,m;
            scanf("%d%d",&n,&m);
            dp[1][1]=m;
            for(int i=2;i<=n;i++)
            {
                for(int j=1;j<=i&&j<=m;j++)
                {
                    dp[i][j]=(dp[i-1][j]*j%mod+dp[i-1][j-1]*(m-j+1)%mod)%mod;
                }
            }
            ll ans=0;
            for(int i=1;i<m;i++)
            {
                ans=(ans+dp[n][i]*quick_pow(m-i,n)%mod)%mod;
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }


    Fighting~
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  • 原文地址:https://www.cnblogs.com/Archger/p/12774697.html
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