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|
#includefor(int i=0;iif(mat[i][k]==0)continue;
for(int j=0;jif(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;i1;
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)
#includeF后缀数组的简单应用，见以前的博客：
hdu 4691