POJ 3387 Crazy Thairs

POJ_3378

    如果设f[i][j]表示第i个数选a[j]时的方案数，那么f[i][j]就等于所有满足a[k]

    此外，在计算中间结果时5000^4是不会超long long的，但是最后合并的时候需要用到高精度加法。

#include
#include<string.h>
#include
#define MAXD 50010
int N, T, node, left[MAXD], right[MAXD], a[MAXD], key[MAXD], size[MAXD];
long long int sum[MAXD], num[MAXD], wa[MAXD], wb[MAXD], *f, *g;
struct BigInteger
{
int a[30];
void init(long long int n)
    {
int i;
for(i = 0; i < 30; i ++)
        {
            a[i] = n % 10;
            n /= 10;
        }
    }
    BigInteger add(BigInteger &b)
    {
int i, s, c = 0;
        BigInteger ans;
for(i = 0; i < 30; i ++)
        {
            s = a[i] + b.a[i] + c;
            ans.a[i] = s % 10;
            c = s / 10;
        }
return ans;
    }
void print()
    {
int i;
for(i = 29; i > 0; i --)
if(a[i])
break;
for(; i >= 0; i --)
            printf("%d", a[i]);
        printf("\n");
    }
};
void leftrotate(int &T)
{
int k = right[T];
    right[T] = left[k];
    left[k] = T;
    size[k] = size[T];
    size[T] = size[left[T]] + size[right[T]] + 1;
    sum[k] = sum[T];
    sum[T] = sum[left[T]] + sum[right[T]] + num[T];
    T = k;
}
void rightrotate(int &T)
{
int k = left[T];
    left[T] = right[k];
    right[k] = T;
    size[k] = size[T];
    size[T] = size[left[T]] + size[right[T]] + 1;
    sum[k] = sum[T];
    sum[T] = sum[left[T]] + sum[right[T]] + num[T];
    T = k;
}
void maintain(int &T, int flag)
{
if(flag == 0)
    {
if(size[left[left[T]]] > size[right[T]])
            rightrotate(T);
else if(size[right[left[T]]] > size[right[T]])
            leftrotate(left[T]), rightrotate(T);
else
return ;
    }
else
    {
if(size[right[right[T]]] > size[left[T]])
            leftrotate(T);
else if(size[left[right[T]]] > size[left[T]])
            rightrotate(right[T]), leftrotate(T);
else
return ;
    }
    maintain(left[T], 0);
    maintain(right[T], 1);
    maintain(T, 0);
    maintain(T, 1);
}
void init()
{
int i;
for(i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
}
long long int  getsum(int &T, int v)
{
if(T == 0)
return 0;
if(v <= key[T])
return getsum(left[T], v);
else
return getsum(right[T], v) + sum[left[T]] + num[T];
}
void newnode(int &T, long long int n, int v)
{
    T = ++ node;
    key[T] = v;
    size[T] = 1;
    sum[T] = num[T] = n;
    left[T] = right[T] = 0;
}
void Insert(int &T, long long int n, int v)
{
if(T == 0)
    {
        newnode(T, n, v);
return ;
    }
    ++ size[T];
    sum[T] += n;
if(v < key[T])
        Insert(left[T], n, v);
else
        Insert(right[T], n, v);
    maintain(T, v >= key[T]);
}
void solve()
{
int i, j, k;
long long int *t;
    BigInteger ans, x;
    f = wa, g = wb;
for(i = 1; i <= N; i ++)
        g[i] = 1;
for(i = 2; i <= 5; i ++)
    {
        T = node = size[0] = left[0] = right[0] = 0;
for(j = 1; j <= N; j ++)
        {
            f[j] = getsum(T, a[j]);
            Insert(T, g[j], a[j]);
        }
        t = f, f = g, g = t;
    }
    ans.init(0);
for(i = 1; i <= N; i ++)
    {
        x.init(g[i]);
        ans = ans.add(x);
    }
    ans.print();
}
int main()
{
while(scanf("%d", &N) == 1)
    {
        init();
        solve();
    }
return 0;
}