poj 1186 分类： poj 2015-08-08 22:28 4人阅读 评论(0) 收藏

似乎是NOI 2001的题目，感觉很有意思。。。

Meet in the Middle （中途相遇法）

看起来O(MN)不可做，方程移项有真相：k1∗xp11+k2∗xp22+k3∗xp33=−(k4∗xp44+k5∗xp55+k6∗xp66)

分别计算 :

k1∗xp11+k2∗xp22+k3∗xp33=W 的解数fL(W)

k4∗xp44+k5∗xp55+k6∗xp66=W 的解数fR(W)

那么 ans=∑fL(x)∗fR(−x)

时间复杂度：O(MN/2)

另外注意使用 std::unique 函数之后的数组中每个值出现的次数与原数组不相同。。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <iostream>
#include <algorithm>

template<class Num>void read(Num &x)
{
char c; int flag = 1;
while((c = getchar()) < '0' || c > '9')
if(c == '-') flag *= -1;
x = c - '0';
while((c = getchar()) >= '0' && c <= '9')
x = (x<<3) + (x<<1) + (c-'0');
x *= flag;
return;
}
template<class Num>void write(Num x)
{
if(x < 0) putchar('-'), x = -x;
static char s[20];int sl = 0;
while(x) s[sl++] = x%10 + '0',x /= 10;
if(!sl) {putchar('0');return;}
while(sl) putchar(s[--sl]);
}
const int maxn = 10, maxm = 155, size = maxm*maxm*maxm;

int n, m, k[maxn], p[maxn];

int val[size], len, vlen, cnt[size];

long long ans = 0;

void count(int rem,int v)
{
if(rem == 0)
{
val[++len] = v;
return;
}

for(int i = 1; i <= m; i++)
{
int cal = k[rem];

for(int j = 1; j <= p[rem]; j++) cal *= i;

count(rem - 1, v + cal);
}
}
#define find_in_val(x) std::lower_bound(val + 1, val + vlen + 1, x) - val
void prework()
{
static int tmp[size];

for(int i = 1; i <= len; i++) tmp[i] = val[i];

std::sort(val + 1, val + len + 1);
vlen = std::unique(val + 1, val + len + 1) - (val + 1);

for(int i = 1; i <= len; i++) cnt[find_in_val(tmp[i])] ++;
}

void dfs(int pos,int v)
{
if(pos > n)
{
int t = find_in_val(-v);
if(val[t] == -v) ans += cnt[t];
return;
}

for(int i = 1; i <= m; i++)
{
int cal = k[pos];

for(int j = 1; j <= p[pos]; j++) cal *= i;

dfs(pos + 1, v + cal);
}
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("1186.in","r",stdin);
freopen("1186.out","w",stdout);
#endif

read(n), read(m);
for(int i = 1; i <= n; i++)
read(k[i]), read(p[i]);

count(n>>1, 0);

prework();

dfs((n>>1) + 1, 0);

write(ans);

#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
#endif
return 0;
}