poj 1186

来源：互联网发布：无实体店网络营业执照编辑：程序博客网时间：2024/06/11 11:22

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) - valvoid 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;}

0 0