zoj 3662 Math Magic(背包)

题意：k个数，和为n，lcm为m，求情况总数。

用dp[i][j][k]表示放了i个数，和为j，lcm为k的情况总数，但是太大，所以要预处理出m的所有约数，以及lcm，还有映射关系。

这样k就缩减到20左右。内存就开的下了。测了下1000 1000 100是差不多10^7次计算。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll long long
#define ull unsigned long long
#define eps 1e-8
#define NMAX 10000000
#define MOD 1000000007
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
#define mp make_pair
template<class T>
inline void scan_d(T &ret)
{
char c;
int flag = 0;
ret=0;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c == '-')
{
flag = 1;
c = getchar();
}
while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
if(flag) ret = -ret;
}
ll dp[101][1001][35];
int ha[35],cnt;
int pos[1001];
int LCM[1001][1001];
void add(ll &a, ll b)
{
a += b;
if(a > MOD) a -= MOD;
}

int gcd(int a, int b)
{
return (b==0) ? a : gcd(b,a%b);
}

int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
// freopen("o.txt","w",stdout);
#endif
int n,m,K;
for(int i = 1; i <= 1000; i++)
for(int j = 1; j <= 1000; j++)
LCM[i][j] = i*j/gcd(i,j);
while(~scanf("%d%d%d",&n,&m,&K))
{
cnt = 1;
for(int i = 1; i*i <= m; i++) if(m%i == 0)
{
if(i*i != m)
{
ha[cnt++] = i;
pos[i] = cnt-1;
ha[cnt++] = m/i;
pos[m/i] = cnt-1;
}
else
{
ha[cnt++] = i;
pos[i] = cnt-1;
}
}
for(int i = 0; i <= K; i++)
for(int j = 0; j <= n; j++)
for(int k = 0; k < cnt; k++) dp[i][j][k] = 0;
dp[0][0][0] = 1;
int tmp = 0;
for(int j = 0; j < K; j++)
for(int k = 0; k <= n; k++)
for(int l = 0; l < cnt; l++) if(dp[j][k][l] != 0)
for(int i = 1; i < cnt; i++) if(k + ha[i] <= n)
{
tmp++;
int lc;
if(l == 0) lc = ha[i];
else lc = LCM[ha[l]][ha[i]];
add(dp[j+1][k+ha[i]][pos[lc]],dp[j][k][l]);
}
printf("%lld\n",dp[K]
[pos[m]]);
}
return 0;
}