2013 Multi-University Training Contest 9（hdu 4686 - 4691）dp（好）+矩阵快速幂+一般图匹配带花树+后缀数组

A - 1001

Description

An Arc of Dream is a curve defined by following function:

AoD(n)=∑n−1i=0ai∗bi

where
a0=A0
ai=ai−1∗AX+AY
b0=B0
bi=bi−1∗BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

Input
There are multiple test cases. Process to the End of File.
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 10 18, and all the other integers are no more than 2×10 9.

Output
For each test case, output AoD(N) modulo 1,000,000,007.

Sample Input

1
1 2 3
4 5 6
2
1 2 3
4 5 6
3
1 2 3
4 5 6

Sample Output

4
134
1902
构造矩阵，其实就是由aibi=(ai−1∗AX+AY)∗(bi−1∗BX+BY)推出来的
|aibi| |AXBXAXBYAYBXAYBY00|
|ai| = |0AX0AY00|
|bi| |00BXBY00|
|1| |000100|
|S| |AXBXAXBYAYBXAYBY11|

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int MOD=1e9+7;LL N;LL A0,AX,AY,B0,BX,BY;struct Matrix{    LL mat[10][10];    int n;    Matrix (int _n=5){        n=_n;        memset(mat,0,sizeof(mat));    }    Matrix operator*(const Matrix &a)const{        Matrix res;        for(int k=0;k<n;k++){            for(int i=0;i<n;i++){                if(mat[i][k]==0)continue;                for(int j=0;j<n;j++){                    if(a.mat[k][j]==0)continue;                    res.mat[i][j]=(res.mat[i][j]+mat[i][k]*a.mat[k][j]%MOD);                    if(res.mat[i][j]>=MOD)res.mat[i][j]-=MOD;                }            }        }        return res;    }};Matrix pow_mul(Matrix A,LL n){    Matrix res;    for(int i=0;i<res.n;i++)res.mat[i][i]=1;    while(n){        if(n&1)res=res*A;        A=A*A;        n>>=1;    }    return res;}int main(){    while(scanf("%I64d",&N)!=EOF){        scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&A0,&AX,&AY,&B0,&BX,&BY);        Matrix A,B;        if(N==0){            printf("0\n");            continue;        }        A0%=MOD,B0%=MOD,AX%=MOD,AY%=MOD,BX%=MOD,BY%=MOD;        A.mat[0][0]=A.mat[4][0]=AX*BX%MOD;        A.mat[0][1]=A.mat[4][1]=AX*BY%MOD;        A.mat[0][2]=A.mat[4][2]=AY*BX%MOD;        A.mat[0][3]=A.mat[4][3]=AY*BY%MOD;A.mat[4][4]=1;        A.mat[1][1]=AX;A.mat[1][3]=AY;        A.mat[2][2]=BX,A.mat[2][3]=BY;        A.mat[3][3]=1;        B.mat[0][0]=B.mat[4][0]=A0*B0%MOD,B.mat[1][0]=A0,B.mat[2][0]=B0;        B.mat[3][0]=1;        A=pow_mul(A,N-1);        A=A*B;        printf("%I64d\n",A.mat[4][0]);    }    return 0;}

B题是一般图匹配，见前面的博客：
hdu - 4687

D - 1004

Description

A derangement is a permutation such that none of the elements appear in their original position. For example, [5, 4, 1, 2, 3] is a derangement of [1, 2, 3, 4, 5]. Subtracting the original permutation from the derangement, we get the derangement difference [4, 2, -2, -2, -2], where none of its elements is zero. Taking the signs of these differences, we get the derangement sign [+, +, -, -, -]. Now given a derangement sign, how many derangements are there satisfying the given derangement sign?

Input

There are multiple test cases. Process to the End of File.
Each test case is a line of derangements sign whose length is between 1 and 20, inclusively.

Output

For each test case, output the number of derangements.

Sample Input

+-
++—

Sample Output

1
13
递推，令dp[i][j]表示错位排列的前i个数中有j个对应+的位置尚未填写。这里的j还有令一层意思，即0~i-1中也恰好有j个数未出现在已知的部分错位排列中。原因是如果某个位置填写了数，那么必须是0~i-1其中之一，而有j个位置尚未填写，故0~i-1中有j个未出现。

假如已经考虑了前i位，考虑a[i]：

+：这个位置的数也无法确定，算作“未填写”，而之前j个未填的地方可以挑一个填i，对应：dp[i+1][j] += dp[i][j] * j；也可以不用i，对应：dp[i+1][j+1] += dp[i][j]-：这个位置的数必须确定，由于0~i-1中恰有j个数未被填写，所以位置i可以填上这j个数中的任何一个。另外之前j个未确定的位置可以填i(对应：dp[i+1][j-1] += dp[i][j] * j * j)；也可以不用i(对应：dp[i+1][j] += dp[i][j] * j)

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=50;typedef long long LL;LL dp[maxn][maxn];int N;char s[maxn];LL DP(int i,int j){    if(dp[i][j]!=-1)return dp[i][j];    if(i==N){        dp[i][j]=(j==0);        return dp[i][j];    }    LL ans=0;    if(s[i]=='+'){        ans+=j*DP(i+1,j);        ans+=DP(i+1,j+1);    } else {        ans+=j*j*DP(i+1,j-1);        ans+=j*DP(i+1,j);    }    return dp[i][j]=ans;}int main(){    while(scanf("%s",s)!=EOF){        N=strlen(s);        memset(dp,-1,sizeof(dp));        printf("%I64d\n",DP(0,0));    }    return 0;}

F后缀数组的简单应用，见以前的博客：
hdu 4691