H - Math Magic
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4427
Description
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 (A 1, A 2, ..., A i, A i+1,..., A K) = N 2. LCM (A 1, A 2, ..., A i, A i+1,..., A K) = M Can you calculate how many kinds of solutions are there for A i (A i 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(10 9 + 7). You can get more details in the sample and hint below.
Sample Input
4 2 23 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).
思路： dp[K][N][M]; K个 ，和，LCM
dp[K][N][M]=sum(dp[K-1][N-x][y]);LCM(x,y)=M;
#include<cstdio>#include<cstring>#include<iostream>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int mm=1002;const int mod=1e9+7;int LCM[mm][mm];int dp[2][mm][mm],ppp[mm],pos;int gcd(int a,int b){ int c; while(b) { c=b;b=a%b;a=c; } return a;}int lcm(int a,int b){ return a*b/gcd(a,b);}int main(){ FOR(i,1,mm-1)FOR(j,1,mm-1) LCM[i][j]=lcm(i,j); int N,M,K;///dp num sum LCM while(~scanf("%d%d%d",&N,&M,&K)) { pos=0; FOR(i,1,M) if(M%i==0) ppp[pos++]=i; int now=0; FOR(i,0,N)FOR(j,0,pos-1) dp[now][i][ ppp[j] ]=0; dp[now][0][1]=1; FOR(i,1,K)///K次 { now^=1; FOR(j,0,N)FOR(k,0,pos-1) dp[now][j][ ppp[k] ]=0;///clear dp FOR(j,i-1,N)///至少是1 FOR(k,0,pos-1) { if(dp[now^1][j][ ppp[k] ]==0)continue; FOR(p,0,pos-1)///新加p { int x=j+ppp[p]; int y=LCM[ ppp[p] ][ ppp[k] ]; if(x>N||M%y!=0)continue; dp[now][x][y]+=dp[now^1][j][ ppp[k] ]; if(dp[now][x][y]>=mod) dp[now][x][y]-=mod; } } } printf("%d\n",dp[now][N][M]); }}