## ZOJ 3662 Math Magic 第37届ACM/ICPC长春赛区H题（DP）

Math Magic

Time Limit: 3 Seconds      Memory Limit: 32768 KB

Yesterday, my teacher taught us about math: +, -, *, /, GCD, LCM... As you know, LCM (Least common multiple) of two positive numbers can be solved easily because of a * b = GCD (a, b) * LCM (a, b).

In class, I raised a new idea: "how to calculate the LCM of K numbers". It's also an easy problem indeed, which only cost me 1 minute to solve it. I raised my hand and told teacher about my outstanding algorithm. Teacher just smiled and smiled...

After class, my teacher gave me a new problem and he wanted me solve it in 1 minute, too. If we know three parameters N, M, K, and two equations:

1. SUM (A1, A2, ..., Ai, Ai+1,..., AK) = N
2. LCM (A1, A2, ..., Ai, Ai+1,..., AK) = M

Can you calculate how many kinds of solutions are there for Ai (Ai are all positive numbers). I began to roll cold sweat but teacher just smiled and smiled.

Can you solve this problem in 1 minute?

#### Input

There are multiple test cases.

Each test case contains three integers N, M, K. (1 ≤ N, M ≤ 1,000, 1 ≤ K ≤ 100)

#### Output

For each test case, output an integer indicating the number of solution modulo 1,000,000,007(1e9 + 7).

You can get more details in the sample and hint below.

#### Sample Input

4 2 2
3 2 2


#### Sample Output

1
2


#### Hint

The first test case: the only solution is (2, 2).

The second test case: the solution are (1, 2) and (2, 1).

Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

dp[now][i][j]表示当前状态下，和为i,LCM为j的解的个数。递推K次就出答案了。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int MOD=1000000007;
int dp[2][1010][1010];

int num[1000];

int gcd(int a,int b)
{
if(b==0)return a;
return gcd(b,a%b);
}
int lcm(int a,int b)
{
return a/gcd(a,b)*b;
}
int LCM[1010][1010];
int main()
{
int n,m,k;
for(int i=1;i<=1000;i++)
for(int j=1;j<=1000;j++)
LCM[i][j]=lcm(i,j);

while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
int cnt=0;
for(int i=1;i<=m;i++)
{
if(m%i==0)
num[cnt++]=i;
}
int now=0;
//memset(dp[now],0,sizeof(dp[now]));
for(int i=0;i<=n;i++)
for(int j=0;j<cnt;j++)
dp[now][i][num[j]]=0;
dp[now][0][1]=1;

for(int t=1;t<=k;t++)
{
now^=1;
// memset(dp[now],0,sizeof(dp[now]));
for(int i=0;i<=n;i++)
for(int j=0;j<cnt;j++)
dp[now][i][num[j]]=0;
for(int i=t-1;i<=n;i++)
for(int j=0;j<cnt;j++)
{
if(dp[now^1][i][num[j]]==0)continue;
for(int p=0;p<cnt;p++)
{
int x=i+num[p];
//int y=lcm(num[j],num[p]);
int y=LCM[num[j]][num[p]];
if(x>n||m%y!=0)continue;
dp[now][x][y]+=dp[now^1][i][num[j]];
dp[now][x][y]%=MOD;
}
}
}
printf("%d\n",dp[now][n][m]);
}
return 0;
}

posted on 2012-10-17 13:35  kuangbin  阅读(1365)  评论(0编辑  收藏  举报

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