Zoj3662math Magic (Group backpack + full backpack)

I-Math MagicTime limit:MS Memory Limit:32768KB 64bit IO Format:%lld & Amp %llu SubmitStatus

Description

Yesterday, my teacher taught us about math: +,-, *,/, GCD, LCM ... As you know, LCM (Least Common multiple) of the 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 both equations:

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

Can calculate how many kinds of solutions is there for AI (AI is 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 is 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 23) 2 2

Sample Output

12

Hint

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

The second Test case:the solution is (1, 2) and (2, 1).

Test instructions: There are k positive integers, and n, least common multiple m, how many combinations.

Analysis: Because this k number is the factor of M, and all the factors of M is the case of all possible least common multiple, however, n,m<=1000, so the maximum number of factors within 1000 is 35, so just record it, with the idea of discretization can be solved.

#include <stdio.h> #include <string.h>const int MOD = 1e9+7;int Dp[2][1005][1005];int lcm[1005][1005],n,m,k    , Num[105],cet;int gcd (int a,int b) {if (b==0) return A; return gcd (b,a%b);} int LCM (int a,int b) {return (A*B)/GCD (A, b);}    int main () {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) {cet=0;        for (int i=1;i<=m;i++) if (m%i==0) num[++cet]=i;        for (int i=0;i<=n;i++)//And for (int j=1;j<=cet;j++)//least common multiple dp[0][i][num[j]]=0;        Dp[0][0][1]=1;        int flag=0,ss,ff;            for (int k=1;k<=k;k++) {flag^=1;            for (int i=k;i<=n;i++) for (int j=1;j<=cet;j++) dp[flag][i][num[j]]=0; for (int s=k-1;s<=n;s++)//k-1 The number of at least k-1 for (int c=1;c<=cet;c++)//least common multiple {if (dp[               flag^1][s][num[c]]==0)     Continue                    for (int i=1;i<=cet;i++)//per factor {Ss=s+num[i];                    Ff=lcm[num[c]][num[i]]; if (ss>n| |                    m%ff!=0) continue;                    DP[FLAG][SS][FF]+=DP[FLAG^1][S][NUM[C]];                Dp[flag][ss][ff]%=mod;    }}} printf ("%d\n", Dp[flag][n][m]); }}

Zoj3662math Magic (Group backpack + full backpack)