2016长乐夏令营 Day10

T1:

f[i][j]:将每个数字转换为二进制数码后前八位为i的数中与后八位为j的数中的后八位数码opt操作后的最大值。对于每个位置pos查询与更新都是根号65536的#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<bitset>
using namespace std;

const int maxn = 1<<9;
const int maxm = 1E5 + 10;
typedef unsigned short UI;
const UI T = 8;

int n,opt,typ,f[maxn][maxn],g[maxn][maxn],ans1[maxm],ans2[maxm];
char ch[10];

int Read()
{
int ret = 0;
char c = getchar();
while (c < '0' || '9' < c) c = getchar();
while ('0' <= c && c <= '9') ret = ret*10 + c - '0',c = getchar();
return ret;
}

int cal(int x,int y,int opt)
{
if (opt == 1) return x & y;
if (opt == 2) return x | y;
return x ^ y;
}

void Work(UI A,UI B,int i)
{
for (int j = 0; j < (1<<8); j++) {
int now = f[j];
now += (cal(j,A,opt)<<8);
if (now > ans1[i]) ans1[i] = now,ans2[i] = g[j][B];
else if (now == ans1[i]) ans2[i] += g[j][B];
}
for (int j = 0; j < (1<<8); j++) {
int now = cal(j,B,opt);
if (f[A][j] < now) f[A][j] = now,g[A][j] = 1;
else if (f[A][j] == now) ++g[A][j];
}
}

int main()
{
#ifdef DMC
freopen("DMC.txt","r",stdin);
freopen("TEST.txt","w",stdout);
#else
freopen("friend.in","r",stdin);
freopen("friend.out","w",stdout);
#endif

n = Read();
for (int i = 0; i < (1<<9); i++)
for (int j = 0; j < (1<<9); j++)
f[i][j] = -(1<<17),g[i][j] = 0;
scanf("%s",ch);
if (ch[0] == 'a') opt = 1;
else if (ch[0] == 'o') opt = 2;
else opt = 3;
typ = Read();
for (int i = 1; i <= n; i++) {
UI x = Read();
UI A = (x>>T);//,B = ((x<<T)>>T);
UI B = x<<T; B >>= T;
Work(A,B,i);
}
for (int i = 2; i <= n; i++) {
if (typ) printf("%d %d\n",ans1[i],ans2[i]);
else printf("%d\n",ans1[i]);
}
return 0;
}



[b]T2：

g[i][j][k]:同种颜色的k辆车占据i行j列(每行都有车)的方案数
首先是C(i*j,k)但是这肯定有重复
用容斥原理
总数 - 至少一行没放车的方案 - 至少一列没放车的方案 + 至少一行一列没放车的方案...
g[i][j][k] = ∑∑(-1)^(p+q)*C(i,p)*C(j,q)*C((i-p)*(j-q),k)

f[i][j][k]:前k种颜色的车占据i行j列的方案数
f[i][j][k] = ∑∑C(i,p)*C(j,q)*f[p][q][k-1]*g[i-p][i-q][a[k]]



那么第二问的答案就是f
[m][k]了

至于第一问[b]∑∑C(n,i)*C(m,j)*f
[m][k][/b]#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<bitset>
using namespace std;

const int maxn = 3001*3000;
typedef long long LL;
const LL mo = 1e9 + 9;

int n,m,k,typ,Max,a[15];
LL ans = 0,f[31][31][11],g[31][31][901],fac[maxn],inv[maxn];

LL C(int n,int m)
{
return fac
*inv[m]%mo*inv[n-m]%mo;
}

LL cal(int i,int j,int l)
{
if (max(i,j) > a[l] || i*j < a[l]) return 0;
LL ret = 0;
for (int p = 0; p <= i; p++)
for (int q = 0; q <= j; q++) {
if ((i-p)*(j-q) < a[l]) continue;
int x = ((p+q)&1)?-1:1;
LL Add = C(i,p)*C(j,q)%mo*C((i-p)*(j-q),a[l])%mo;
if (x == -1) ret = (ret - Add + mo)%mo;
else ret = (ret + Add) % mo;
}
return ret;
}

LL ksm(LL x,LL y)
{
LL ret = 1;
for (; y; y >>= 1) {
if (y&1) ret = ret*x%mo;
x = x*x%mo;
}
return ret;
}

int main()
{
#ifdef DMC
freopen("DMC.txt","r",stdin);
//freopen("test1.txt","w",stdout);
#else
freopen("chessboard.in","r",stdin);
freopen("chessboard.out","w",stdout);
#endif

cin >> n >> m >> k;
fac[0] = 1; inv[0] = 1;
for (LL i = 1; i <= n*m; i++) fac[i] = fac[i-1]*i%mo;
inv[n*m] = ksm(fac[n*m],mo-2);
for (LL i = n*m; i >= 1; i--)
inv[i-1] = inv[i]*i%mo;
for (int i = 1; i <= k; i++) scanf("%d",&a[i]),Max = max(Max,a[i]);
cin >> typ; //g[0][0][0] =
f[0][0][0] = 1;
//LL CA = cal(12,28,1);
if (n > 30 || m > 30) {
ans = cal(n,m,k);
cout << ans;
}
else {
for (int l = 1; l <= k; l++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
g[i][j][l] = cal(i,j,l);
for (int l = 1; l <= k; l++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int p = 0; p <= i; p++)
for (int q = 0; q <= j; q++)
f[i][j][l] = (f[i][j][l] + C(i,p)*C(j,q)%mo*f[p][q][l-1]%mo*g[i-p][j-q][l]%mo)%mo;
if (typ) cout << f
[m][k] << endl;
else {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
ans = (ans + f[i][j][k]*C(n,i)%mo*C(m,j)%mo)%mo;
cout << ans << endl;
}
}
return 0;
}

[b]为了方便算组合数，预处理阶乘和逆元[/b]

[b]fac的逆元有快速计算方式:[/b]

[b]inv[n-1] = inv
*n[/b]



[b]T3:[/b]

[b]提答。。万恶的提答[/b]